This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the
length of each side, then there are 3 unknowns. So here\'s what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.
Solution
This word problem is about a triangle whose perimeter is 47 miles. Since it\'s asking to find the
length of each side, then there are 3 unknowns. So here\'s what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here\'s how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined l.
Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
Module 5 (Part 3-Revised)-Functions and Relations.pdfGaleJean
The Cavite Mutiny of 1872 was an uprising of military personnel of Fort San Felipe, the Spanish arsenal in Cavite, Philippines on January 20, 1872, about 200 Filipino military personnel of Fort San Felipe Arsenal in Cavite, Philippines, staged a mutiny which in a way led to the Philippine Revolution in 1896. The 1872 Cavite Mutiny was precipitated by the removal of long-standing personal benefits to the workers such as tax (tribute) and forced labor exemptions on order from the Governor General Rafael de Izquierdo. Many scholars believe that the Cavite Mutiny of 1872 was the beginning of Filipino nationalism that would eventually lead to the Philippine Revolution of 1896.
Mathematics (from Greek μάθημα máthēma, “knowledge, study, learning”) is the study of topics such as quantity (numbers), structure, space, and change. There is a range of views among mathematicians and philosophers as to the exact scope and definition of mathematics
Module 5 (Part 3-Revised)-Functions and Relations.pdfGaleJean
The Cavite Mutiny of 1872 was an uprising of military personnel of Fort San Felipe, the Spanish arsenal in Cavite, Philippines on January 20, 1872, about 200 Filipino military personnel of Fort San Felipe Arsenal in Cavite, Philippines, staged a mutiny which in a way led to the Philippine Revolution in 1896. The 1872 Cavite Mutiny was precipitated by the removal of long-standing personal benefits to the workers such as tax (tribute) and forced labor exemptions on order from the Governor General Rafael de Izquierdo. Many scholars believe that the Cavite Mutiny of 1872 was the beginning of Filipino nationalism that would eventually lead to the Philippine Revolution of 1896.
Engineering Maths N4
Rectangular Form
Polar Form
Relationship between Polar and Rectangular Complex NUmbers
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November 2015 Exams Registrations
Enrolment for Fet College Exams
Distance Learning
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Extra Lessons
E Learning
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Phone 073 090 2954
Fax 086 244 2355
Email topstudentz017@gmail.com
An ellipse is the locus of a point which moves in such a way that its distance form a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix and the constant ratio is called the eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of a point which moves in a plane so that the sum of it distances from fixed points is constant.
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a (fixed distance) then
(ii) Major & Minor axis : The straight line AA is called major axis and BB is called minor axis. The major and minor axis taken together are called the principal axes and its length will be given by
Length of major axis 2a Length of minor axis 2b
(iii) Centre : The point which bisect each chord of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci of an ellipse.
(vi) Latus Rectum : Such chord which passes through either focus and perpendicular to the major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae, 𝑙 ) then 2𝑙 is the length of
SP+S'P=
{(h ae)2 k 2} +
= 2a
Latus Rectum.
Length of Latus rectum is given by
2b2
.
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by. x2(1– e2) + y2 = a2(1– e2)
2
a
(vii) Relation between constant a, b, and e
a 2 b2
b2 = a2(1– e2) e2 =
a 2
x2
a 2 +
y
a 2 (1 e 2 ) = 1
e =
a 2
Result :
Major Axis
(a) Centre C is the point of intersection of the axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x 2 y2
a 2 b2
1 when a < b.
Directrix Minor Axis Directrix x = -a/e x = a/e
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2 y2
a2 b2
2.1.1 Various parameter related with standard ellipse :
In this case major axis is BB= 2b which is along y- axis and minor axis is AA= 2a along x- axis. Focus S(0,be) and S(0,–be) and directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is (h,k) and the directrix is the line ax + by + c = 0 and the eccentricity will be e. Then let P(x1,y1) be any point on the ellipse which moves such that SP = ePM
Let the equation of the ellipse x
y2
a > b
(x –h)2 + (y –k)2 =
e 2 (ax1 by1 c) 2
a 2 b2
1 1 a 2 b2
(i) Vertices of an ellipse : The point of which ellipse cut the axis x-axis at (a,0) & (– a, 0) and y- axis at (0, b) & (0, – b) is called the vertices of an ellipse.
Hence the locus of (x1,y1) will be given by (a2 + b
Solution of a Sangaku ``Tangency" Problem via Geometric AlgebraJames Smith
Because the shortage of worked-out examples at introductory levels is an obstacle to widespread adoption of Geometric Algebra (GA), we use GA to solve one of the beautiful \emph{sangaku} problems from 19th-Century Japan. Among the GA operations that prove useful is the rotation of vectors via the unit bivector
The metal will undergo oxidation forming metal ion and releasing .pdfKUNALHARCHANDANI1
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas]
Solution
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas].
Static Keyword Static is a keyword in C++ used to give special chara.pdfKUNALHARCHANDANI1
Static Keyword Static is a keyword in C++ used to give special characteristics to an element.
Static elements are allocated storage only once in a program lifetime in static storage area. And
they have a scope till the program lifetime. Static Keyword can be used with following, Static
variable in functions Static Class Objects Static member Variable in class Static Methods in class
Static variables inside Functions Static variables when used inside function are initialized only
once, and then they hold there value even through function calls. These static variables are stored
on static storage area , not in stack. void counter() { static int count=0; cout << count++; } int
main(0 { for(int i=0;i<5;i++) { counter(); } } Output : 0 1 2 3 4 Let\'s se the same program\'s
output without using static variable. void counter() { int count=0; cout << count++; } int main(0
{ for(int i=0;i<5;i++) { counter(); } } Output : 0 0 0 0 0 If we do not use static keyword, the
variable count, is reinitialized everytime when counter() function is called, and gets destroyed
each time when counter() functions ends. But, if we make it static, once initialized count will
have a scope till the end of main() function and it will carry its value through function calls too.
If you don\'t initialize a static variable, they are by default initialized to zero. Static class Objects
Static keyword works in the same way for class objects too. Objects declared static are allocated
storage in static storage area, and have scope till the end of program. Static objects are also
initialized using constructors like other normal objects. Assignment to zero, on using static
keyword is only for primitive datatypes, not for user defined datatypes. class Abc { int i; public:
Abc() { i=0; cout << \"constructor\"; } ~Abc() { cout << \"destructor\"; } }; void f() { static Abc
obj; } int main() { int x=0; if(x==0) { f(); } cout << \"END\"; } Output : constructor END
destructor You must be thinking, why was destructor not called upon the end of the scope of if
condition. This is because object was static, which has scope till the program lifetime, hence
destructor for this object was called when main() exits. Static data member in class Static data
members of class are those members which are shared by all the objects. Static data member has
a single piece of storage, and is not available as separate copy with each object, like other non-
static data members. Static member variables (data members) are not initialied using constructor,
because these are not dependent on object initialization. Also, it must be initialized explicitly,
always outside the class. If not initialized, Linker will give error. class X { static int i; public:
X(){}; }; int X::i=1; int main() { X obj; cout << obj.i; // prints value of i } Once the definition for
static data member is made, user cannot redefine it. Though, arithmetic operations can be
performed on it. Static Member Functions These functions work for the .
More Related Content
Similar to This word problem is about a triangle whose perimeter is 47 miles. S.pdf
Engineering Maths N4
Rectangular Form
Polar Form
Relationship between Polar and Rectangular Complex NUmbers
FET College Registrations in Johannesburg Cbd
November 2015 Exams Registrations
Enrolment for Fet College Exams
Distance Learning
Correspondence
Unisa Tutorials
Extra Lessons
E Learning
Engineering Certificates
Engineering Diplomas
Business Certificates
Phone 073 090 2954
Fax 086 244 2355
Email topstudentz017@gmail.com
An ellipse is the locus of a point which moves in such a way that its distance form a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix and the constant ratio is called the eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of a point which moves in a plane so that the sum of it distances from fixed points is constant.
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a (fixed distance) then
(ii) Major & Minor axis : The straight line AA is called major axis and BB is called minor axis. The major and minor axis taken together are called the principal axes and its length will be given by
Length of major axis 2a Length of minor axis 2b
(iii) Centre : The point which bisect each chord of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci of an ellipse.
(vi) Latus Rectum : Such chord which passes through either focus and perpendicular to the major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae, 𝑙 ) then 2𝑙 is the length of
SP+S'P=
{(h ae)2 k 2} +
= 2a
Latus Rectum.
Length of Latus rectum is given by
2b2
.
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by. x2(1– e2) + y2 = a2(1– e2)
2
a
(vii) Relation between constant a, b, and e
a 2 b2
b2 = a2(1– e2) e2 =
a 2
x2
a 2 +
y
a 2 (1 e 2 ) = 1
e =
a 2
Result :
Major Axis
(a) Centre C is the point of intersection of the axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x 2 y2
a 2 b2
1 when a < b.
Directrix Minor Axis Directrix x = -a/e x = a/e
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2 y2
a2 b2
2.1.1 Various parameter related with standard ellipse :
In this case major axis is BB= 2b which is along y- axis and minor axis is AA= 2a along x- axis. Focus S(0,be) and S(0,–be) and directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is (h,k) and the directrix is the line ax + by + c = 0 and the eccentricity will be e. Then let P(x1,y1) be any point on the ellipse which moves such that SP = ePM
Let the equation of the ellipse x
y2
a > b
(x –h)2 + (y –k)2 =
e 2 (ax1 by1 c) 2
a 2 b2
1 1 a 2 b2
(i) Vertices of an ellipse : The point of which ellipse cut the axis x-axis at (a,0) & (– a, 0) and y- axis at (0, b) & (0, – b) is called the vertices of an ellipse.
Hence the locus of (x1,y1) will be given by (a2 + b
Solution of a Sangaku ``Tangency" Problem via Geometric AlgebraJames Smith
Because the shortage of worked-out examples at introductory levels is an obstacle to widespread adoption of Geometric Algebra (GA), we use GA to solve one of the beautiful \emph{sangaku} problems from 19th-Century Japan. Among the GA operations that prove useful is the rotation of vectors via the unit bivector
The metal will undergo oxidation forming metal ion and releasing .pdfKUNALHARCHANDANI1
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas]
Solution
The metal will undergo oxidation forming metal ion and releasing electron
K ==============> K+ + e-
HCl will dissociate into
HCl =============> H+ + Cl-
This H+ will get reduced to hydrogen gas by taking the electron released by the metal.
H+ + e- =========> H2 (gas)
Then the metal ion will combine with chloride forming metal chloride
K+ + Cl- ====> Kcl
The over all reaction would be
K + HCl =====> Kcl + H2 [not balanced]
Note: when metal react with acid, it will form metal salt releasing hydrogen gas].
Static Keyword Static is a keyword in C++ used to give special chara.pdfKUNALHARCHANDANI1
Static Keyword Static is a keyword in C++ used to give special characteristics to an element.
Static elements are allocated storage only once in a program lifetime in static storage area. And
they have a scope till the program lifetime. Static Keyword can be used with following, Static
variable in functions Static Class Objects Static member Variable in class Static Methods in class
Static variables inside Functions Static variables when used inside function are initialized only
once, and then they hold there value even through function calls. These static variables are stored
on static storage area , not in stack. void counter() { static int count=0; cout << count++; } int
main(0 { for(int i=0;i<5;i++) { counter(); } } Output : 0 1 2 3 4 Let\'s se the same program\'s
output without using static variable. void counter() { int count=0; cout << count++; } int main(0
{ for(int i=0;i<5;i++) { counter(); } } Output : 0 0 0 0 0 If we do not use static keyword, the
variable count, is reinitialized everytime when counter() function is called, and gets destroyed
each time when counter() functions ends. But, if we make it static, once initialized count will
have a scope till the end of main() function and it will carry its value through function calls too.
If you don\'t initialize a static variable, they are by default initialized to zero. Static class Objects
Static keyword works in the same way for class objects too. Objects declared static are allocated
storage in static storage area, and have scope till the end of program. Static objects are also
initialized using constructors like other normal objects. Assignment to zero, on using static
keyword is only for primitive datatypes, not for user defined datatypes. class Abc { int i; public:
Abc() { i=0; cout << \"constructor\"; } ~Abc() { cout << \"destructor\"; } }; void f() { static Abc
obj; } int main() { int x=0; if(x==0) { f(); } cout << \"END\"; } Output : constructor END
destructor You must be thinking, why was destructor not called upon the end of the scope of if
condition. This is because object was static, which has scope till the program lifetime, hence
destructor for this object was called when main() exits. Static data member in class Static data
members of class are those members which are shared by all the objects. Static data member has
a single piece of storage, and is not available as separate copy with each object, like other non-
static data members. Static member variables (data members) are not initialied using constructor,
because these are not dependent on object initialization. Also, it must be initialized explicitly,
always outside the class. If not initialized, Linker will give error. class X { static int i; public:
X(){}; }; int X::i=1; int main() { X obj; cout << obj.i; // prints value of i } Once the definition for
static data member is made, user cannot redefine it. Though, arithmetic operations can be
performed on it. Static Member Functions These functions work for the .
Sr2+ is most likely to substitute for Ca2+ becaus.pdfKUNALHARCHANDANI1
Sr2+ is most likely to substitute for Ca2+ because they are in the same group in
Periodic table and has most common properties. Good Luck! Please rate.
Solution
Sr2+ is most likely to substitute for Ca2+ because they are in the same group in
Periodic table and has most common properties. Good Luck! Please rate..
There should only have one singlet resonance for .pdfKUNALHARCHANDANI1
There should only have one singlet resonance for 1,1,2,2-tetrachloroethane. There is
one hydrogen on each carbon. both of them have the exact same environment. Thus, it is shown
as 1 singlet resonance.
Solution
There should only have one singlet resonance for 1,1,2,2-tetrachloroethane. There is
one hydrogen on each carbon. both of them have the exact same environment. Thus, it is shown
as 1 singlet resonance..
the link is not working can u pls write questions.pdfKUNALHARCHANDANI1
the link is not working can u pls write questions dirctly here itself thanks
Solution
the link is not working can u pls write questions dirctly here itself thanks.
The one have higher value of E(cell) is acting as.pdfKUNALHARCHANDANI1
The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e---
>2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode
= 1.07- (-0.13) =1.20V
Solution
The one have higher value of E(cell) is acting as cathode. So, Cathode : Br2 +2e---
>2Br- E= +1.07v Anode : Pb+2 +2e--->Pb E= -0.13v E(cell) = Ecathode - Eanode
= 1.07- (-0.13) =1.20V.
HClO4 in aqueous medium ionizes as Hydrogen(+1)ca.pdfKUNALHARCHANDANI1
HClO4 in aqueous medium ionizes as Hydrogen(+1)cation and Chlorate (-1) anion
Molar conc of Hydrogen cation =.245 Molar conc of chlorate anion = .245
Solution
HClO4 in aqueous medium ionizes as Hydrogen(+1)cation and Chlorate (-1) anion
Molar conc of Hydrogen cation =.245 Molar conc of chlorate anion = .245.
Which of the following would be a description of a system unitA c.pdfKUNALHARCHANDANI1
Which of the following would be a description of a system unit?
A case and internal components used to process data into information
A computer’s main memory component
The module that interprets and carries out computer instructions
A data and information storage device
Ans: A case and internal components used to process data into information
Question 2
Select one answer.
True or false? A tower or minitower PC is a type of all-in-one desktop unit.
True
False
Ans: True
Question 3
Fill in the blanks by selecting one option from each menu.
A/an --expansion card internal drive bay processor chip is inserted on the motherboard and
provides more memory or controls a peripheral device such as a graphics adapter that supports
video games.
Ans: Expansion card
Question 4
Select one answer
What is a peripheral?
A high-speed storage location that temporarily holds data and instructions for future use
A component of the system board that controls the timing of computer activities
A slot on the motherboard that holds memory modules and helps speed computer processes
A device that connects to the system unit and is controlled by the computer’s processor
Ans: A device that connects to the system unit and is controlled by the computer’s processor
Question 5
Fill in the blanks by selecting one option from each menu.
If you often play the latest computer games on your PC, which of the following would you
consider critical components of the PC? Select Yes or No for each option.
The weight and size of your PC
--YesNo
Ans: No
Sophisticated graphics and sound adapter cards
--YesNo
Ans: Yes
Processor speed and amount of RAM
--YesNo
Ans: Yes
Question 1
Select one answer.
If you use your desktop computer only for doing schoolwork, which input and output device
combination would be best from the options below (considering the best combination of cost,
screen resolution, and keyboard design)?
A gaming keyboard and a 27\" LCD monitor
A foldable keyboard and a 15\" LCD monitor
A wireless keyboard and a 24\" LED monitor
A split-style keyboard and a 30\" LED monitor
Ans: A wireless keyboard and a 24\" LED monitor
Question 2
Select one answer.
You are considering using cloud storage as your primary system for saving your data. What
advantages does this storage media have over using your system’s hard drive or some other
external media (external hard drive, flash drive, R/W disks) for storing your data?
You can always get to your data from anywhere.
The data is more secure in cloud storage than on your own storage devices.
The cloud storage provider ensures that you have a backup of your data.
Cloud storage is definitely less expensive than the cost of a hard drive or external devices.
Ans: You can always get to your data from anywhere.
Question 3
Select one answer.
Which of the following can be considered an output device?
Microphone
Stylus
Wireless mouse
Speaker
Ans: Speaker
Question 4
Select one answer.
In this diagram of a simple home network, what appears to be missing?
.
Water is a polar inorganic solvent. Benzene is a nonpolar organic so.pdfKUNALHARCHANDANI1
Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will
generally dissolve solutes with similar polarity (with certain exceptions).
CaCO3 - insoluble in water, insoluble in benzene
CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
NaNO3 - soluble in water, insoluble in benzene
NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
K2SO4 - soluble in water, insoluble in benzene
K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
Petroleum jelly - insoluble in water, soluble in benzene
Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but
soluble in benzene.
Butane - insoluble in water, soluble in benzene
Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene.
Acetone - soluble in water, soluble in benzene
Acetone is a polar organic compound. Thus it is soluble in both water and benzene.
Solution
Water is a polar inorganic solvent. Benzene is a nonpolar organic solvent. The solvents will
generally dissolve solutes with similar polarity (with certain exceptions).
CaCO3 - insoluble in water, insoluble in benzene
CaCO3 is a polar ionic compound that is insoluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
NaNO3 - soluble in water, insoluble in benzene
NaNO3 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
K2SO4 - soluble in water, insoluble in benzene
K2SO4 is a polar ionic compound that is soluble in water. Ionic compounds are insoluble in
nonpolar organic solvents like benzene.
Petroleum jelly - insoluble in water, soluble in benzene
Petroleum jelly is a paraffin and a nonpolar organic compound. Thus it is insoluble in water but
soluble in benzene.
Butane - insoluble in water, soluble in benzene
Butane is a nonpolar organic compound. Thus it is insoluble in water but soluble in benzene.
Acetone - soluble in water, soluble in benzene
Acetone is a polar organic compound. Thus it is soluble in both water and benzene..
viruses which have single strande DNA in their genome come under cat.pdfKUNALHARCHANDANI1
viruses which have single strande DNA in their genome come under category of paroviridae
which generally includes animal pathogens. These viruses when infect the host, the single
stranded DNA is first converted into double stranded DNA, and then using the host machinary
for transcription, ds DNA transcribes into viral mRNA.
For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host
and divided into initiation, elongation and termaination. For gene expression in viruses
transcription is generally not coupled with translation in ssDNA virus. The virus instructs
transcription machinary to make viral components in the host cell.
Solution
viruses which have single strande DNA in their genome come under category of paroviridae
which generally includes animal pathogens. These viruses when infect the host, the single
stranded DNA is first converted into double stranded DNA, and then using the host machinary
for transcription, ds DNA transcribes into viral mRNA.
For transcription, RNA polymerase, transcription factors , enzymes are acquired from the host
and divided into initiation, elongation and termaination. For gene expression in viruses
transcription is generally not coupled with translation in ssDNA virus. The virus instructs
transcription machinary to make viral components in the host cell..
CH4 + 2O2 --> CO2 + 2H2O note: CH4 combustion is a process to react with O2.
Solution
CH4 + 2O2 --> CO2 + 2H2O note: CH4 combustion is a process to react with O2..
Carbon 2 is where D and L differ for all sugars. .pdfKUNALHARCHANDANI1
Carbon 2 is where D and L differ for all sugars. For D and L glucose they will have
the same chirality everywhere cept at 2.
Solution
Carbon 2 is where D and L differ for all sugars. For D and L glucose they will have
the same chirality everywhere cept at 2..
Program to print the Diamond Shape -#include stdio.h int ma.pdfKUNALHARCHANDANI1
Program to print the Diamond Shape :-
#include
int main()
{
int i, j, n;
printf(\"Enter a positive int number greater than one to print the Diamond Shape : \");
scanf(\"%d\", &n);
//To Print the upper pyramid
for(i=1; i<=n; i++)
{
for(j=i; j=1; i--)
{
for(j=i; j<=n; j++)
{
printf(\" \");
}
for(j=2; j<(2*i-1); j++)
{
printf(\"*\");
}
printf(\"\ \");
}
return 0;
}
Solution
Program to print the Diamond Shape :-
#include
int main()
{
int i, j, n;
printf(\"Enter a positive int number greater than one to print the Diamond Shape : \");
scanf(\"%d\", &n);
//To Print the upper pyramid
for(i=1; i<=n; i++)
{
for(j=i; j=1; i--)
{
for(j=i; j<=n; j++)
{
printf(\" \");
}
for(j=2; j<(2*i-1); j++)
{
printf(\"*\");
}
printf(\"\ \");
}
return 0;
}.
Part D option 4 is answerUnless untill they are exposed by some me.pdfKUNALHARCHANDANI1
Part D option 4 is answer
Unless untill they are exposed by some means we can not find them
PArt E option 1 is answer as they hypo... teraposed in rocks
Solution
Part D option 4 is answer
Unless untill they are exposed by some means we can not find them
PArt E option 1 is answer as they hypo... teraposed in rocks.
Isomers which have their atoms connected in the same sequence but di.pdfKUNALHARCHANDANI1
Isomers which have their atoms connected in the same sequence but differ in the way the atoms
are oriented in space - i.e. the difference between two stereoisomers lies only in the three
dimensional arrangement of atoms.
since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and
arrangement is same in 2D space , they are stereo isomers
Solution
Isomers which have their atoms connected in the same sequence but differ in the way the atoms
are oriented in space - i.e. the difference between two stereoisomers lies only in the three
dimensional arrangement of atoms.
since Cis-3-hexene to trans-3-hexene differs in three dimensional arrangement of atoms and
arrangement is same in 2D space , they are stereo isomers.
import java.util.Scanner;
import java.text.DecimalFormat;
import java.io.*;
public class SavingAccount //MUST match the file name!
{
public static void main(String[] args)throws IOException
{
//Create a decimal format for displaying dollars
DecimalFormat dollar = new DecimalFormat(\"#,###.##\");
//Constants
double depositSum = 0.0;
double withdrawalSum = 0.0;
double earnedInt = 0.0;
double startBalance = 500; //Starting balance
// Executables
System.out.println(\"This program solves Programming Challenge 6.11\");
System.out.println();
//Create Scanner object for keyboard input.
Scanner keyboard = new Scanner(System.in);
//Get interest rate
System.out.print(\"Enter the annual interest: \");
double testInterest = keyboard.nextDouble();
//Create an object that accept the starting balance and annual interest
SavingAccount1 account = new SavingAccount1(startBalance, testInterest);
//Open Deposit file.
File file = new File (\"BankDeposits.txt\");
Scanner inputFile = new Scanner(file);
//Read line in file
while (inputFile.hasNext());
{
//Read numbers
double num = inputFile.nextDouble();
//Add the numbers
depositSum += num;
}
//Deposit the file input.
account.deposit(depositSum);
//Close the file
inputFile.close();
//Open Withdrawal file
File file2 = new File(\"BankWithdrawal.txt\");
Scanner inputFile2 = new Scanner(file2);
//Read lines in file
while (inputFile2.hasNext());
{
//Read numbers
double num2 = inputFile2.nextDouble();
//Add the numbers
withdrawalSum += num2;
}
//Withdrawal the file input from account.
account.withdraw(withdrawalSum);
//Close the file
inputFile2.close();
//Add the monthly interest
account.addInt();
//Get amount of interest earned.
earnedInt += account.getInterest();
//Display the data
System.out.println(\"Account balance $\" + dollar.format(account.getBalance()));
System.out.println(\"Total interest earned $\" + dollar.format(account.getInterest()));
}
}//end class
public class SavingAccount1 //MUST match the file name!
{
public static void main(String[] args)
{
System.out.println(\"This program solves Programming Challenge 6.10\");
System.out.println();
}
//Fields
private double balance; //Account balance
private double annualInterest; //annual interest
private double monthInt; //monthly interest
private double earnedInt; //earned interest
double totalWithdraw;
double totalDeposit;
/**
018
This constructor sets the starting balance
019
and the annual interest at 0.0.
020
*/
public void Ward_Tassinda_SavingAccount1()
{
balance = 0.0;
annualInterest = 0.0;
}
/**
029
This constructor set the starting balance and the annual interest rate
030
to the value passed as an argument.
031
@param startBalance The starting balance.
032
*/
public void Ward_Tassinda_SavingAccount1(double startBalance, double interestRate)
{
balance = startBalance;
annualInterest = interestRate;
}
/**
041
This constructor sets the starting balance to
042
the value in the String argument.
043
@param str The starting balance, as a String.
044
*/
public void .
Introduction to AI for Nonprofits with Tapp NetworkTechSoup
Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...Levi Shapiro
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Palestine last event orientationfvgnh .pptxRaedMohamed3
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
This word problem is about a triangle whose perimeter is 47 miles. S.pdf
1. This word problem is about a triangle whose perimeter is 47 miles. Since it's asking to find the
length of each side, then there are 3 unknowns. So here's what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here's how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.
Solution
This word problem is about a triangle whose perimeter is 47 miles. Since it's asking to find the
2. length of each side, then there are 3 unknowns. So here's what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here's how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.
![This word problem is about a triangle whose perimeter is 47 miles. Since it's asking to find the
length of each side, then there are 3 unknowns. So here's what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here's how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.
Solution
This word problem is about a triangle whose perimeter is 47 miles. Since it's asking to find the](https://image.slidesharecdn.com/thiswordproblemisaboutatrianglewhoseperimeteris47miles-230409045234-bb4af606/85/This-word-problem-is-about-a-triangle-whose-perimeter-is-47-miles-S-pdf-1-320.jpg)
![length of each side, then there are 3 unknowns. So here's what we do:
Let the variables S1 be the length of side 1 in miles,
S2 be the length of side 2 in miles,
and S3 be the length of side 3 in miles.
Now, the next thing to do with these variables is to set up equations that follow the word
problem. Here they are as follows:
S1+S2+S3=47 the perimeter of the triangle by adding the lengths of all three sides
S1=(2*S2)-5 1st side (represented by S1) is (represented by =) 5 less (represented by -5) than
twice the 2nd side (represented by 2*S2).
S3=S1+2 3rd side (represented by S3) is 2 more (represented by +2) than the 1st side.
Next, since we have more than one unknown, we should express one equation in terms of one
variable. Since only one variable has a coefficient which is S2 in the term, 2*S2, I suggest that
we express the other variable equations of S1 and S3 in terms of S2. Here's how it goes:
S1=(2*S2)-5 as S1 is already expressed in terms of S2
S3=S1+2=[(2*S2)-5]+2 I plugged in the S1 equation to express S3 in terms of S2.
=(2*S2)-5+2 I combined like terms.
=(2*S2)-3.
Now that I have expressed both S1 and S3 in terms of S2, I can plug those equations into that for
the triangle perimeter:
[(2*S2)-5]+S2+[(2*S2)-3]=47
(2*S2)-5+S2+(2*S2)-3=47 Again, I combined like terms.
(5*S2)-8=47
+8=+8 I added 8 on both sides of the equation to eliminate -8 on the left side.
5*S2=55
5*S2/5=55/5
S2=11
S1=(2*11)-5=22-5=17 Now that I have found the value of S2, I can plug it into the S1 equation.
S3=17+2=19 Now that I have found the value of S1, I can plug it into the S3 equation.
So now here are the lengths of each side of the triangle. Side 1 is 17 miles, side 2 is 11 miles, and
side 3 is 19 miles.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)