1. It will be absorbed by plant. Or remain in soil that .pdf
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mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% = 50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g Solution mass percentage of O in talc = [153.6 / (58.35 + 89.88 + 153.6 + 1.61)] X 100% = 50.6 % mass of O = mass percentage of O x mass of talc = 50.6% x 72.5 = 36.7 g.
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1) Photosynthesis by acquatic plants, algae, seaweed that occurs bel.pdf
1) Photosynthesis by acquatic plants, algae, seaweed that occurs below the surface of water
produces O2 that remians dissolved in the water body.
2) Large quantities of CO2 gets dissolved in the water. This dissolved CO2 reacts with water
molecules to produce O2 as the byproduct. This O2, that is produced as a byproduct, also
remains dissolved in water.
3) O2 also dissolves into the water naturally or with waves or where the water body forms
waterfalls, rapids, etc. The upwelling/downwelling under the water surface helps dissolve O2.
4) Manmade sources of introducing O2 into a water body includes aeration by means of an
airpump. it can be done in a manmade acquatic ecosystems like Aquariums. While, natural water
bodies like ponds can also be rejuvenated using aeration.
Solution
1) Photosynthesis by acquatic plants, algae, seaweed that occurs below the surface of water
produces O2 that remians dissolved in the water body.
2) Large quantities of CO2 gets dissolved in the water. This dissolved CO2 reacts with water
molecules to produce O2 as the byproduct. This O2, that is produced as a byproduct, also
remains dissolved in water.
3) O2 also dissolves into the water naturally or with waves or where the water body forms
waterfalls, rapids, etc. The upwelling/downwelling under the water surface helps dissolve O2.
4) Manmade sources of introducing O2 into a water body includes aeration by means of an
airpump. it can be done in a manmade acquatic ecosystems like Aquariums. While, natural water
bodies like ponds can also be rejuvenated using aeration..
The process of eutrophication occurs in steps- Place the following st.pdf
The process of eutrophication occurs in steps. Place the following steps in their correct order.
1. Algae bloom and put underwater plants in shade, depriving them of sunlight.
2. Plants and algae are consumed by bacteria in the water.
3. Underwater plants die.
4. Nutrients accumulate in the water in excess due to runoff.
a. 3 - 4 - 1 - 2
b. 1 - 4 - 3 - 2
c. 4 - 1 - 3 - 2
d. 1 - 4 - 2 - 3
2. Which of the following is correctly stated?
a. Cultural eutrophication can lead to hypoxia in aquatic systems.
b. Hypoxia is the most common cause of eutrophication, but not cultural eutrophication.
c. Hypoxia is a condition in which aquatic organisms produce too much oxygen.
d. Hypoxia causes low dissolved oxygen concentrations to occur.
3. An ultraoligotrophic lake has been isolated and unchanged for more than a thousand years.
Since we have the ability to grow algae and turn it into oil that we can use to make gasoline, one
day a company buys this property and tries to grow algae in the lake. Given the lack of nutrients,
the algae just won't grow and the company goes out of business. Which statement below about
this lake is correct?
a. The lake has high water quality.
b. The lake has undergone eutrophication
c. This question can't be answered without measuring the concentration of specific nutrients.
d. The lake has low water quality.
Kindly answer the 3 questions. I'll give you an upvote please.
.
Structure Hydra has a tubular body, with only one opening the mouth.pdf
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g
Solution
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g.
Recommended
1) Photosynthesis by acquatic plants, algae, seaweed that occurs bel.pdf
1) Photosynthesis by acquatic plants, algae, seaweed that occurs below the surface of water
produces O2 that remians dissolved in the water body.
2) Large quantities of CO2 gets dissolved in the water. This dissolved CO2 reacts with water
molecules to produce O2 as the byproduct. This O2, that is produced as a byproduct, also
remains dissolved in water.
3) O2 also dissolves into the water naturally or with waves or where the water body forms
waterfalls, rapids, etc. The upwelling/downwelling under the water surface helps dissolve O2.
4) Manmade sources of introducing O2 into a water body includes aeration by means of an
airpump. it can be done in a manmade acquatic ecosystems like Aquariums. While, natural water
bodies like ponds can also be rejuvenated using aeration.
Solution
1) Photosynthesis by acquatic plants, algae, seaweed that occurs below the surface of water
produces O2 that remians dissolved in the water body.
2) Large quantities of CO2 gets dissolved in the water. This dissolved CO2 reacts with water
molecules to produce O2 as the byproduct. This O2, that is produced as a byproduct, also
remains dissolved in water.
3) O2 also dissolves into the water naturally or with waves or where the water body forms
waterfalls, rapids, etc. The upwelling/downwelling under the water surface helps dissolve O2.
4) Manmade sources of introducing O2 into a water body includes aeration by means of an
airpump. it can be done in a manmade acquatic ecosystems like Aquariums. While, natural water
bodies like ponds can also be rejuvenated using aeration..
The process of eutrophication occurs in steps- Place the following st.pdf
The process of eutrophication occurs in steps. Place the following steps in their correct order.
1. Algae bloom and put underwater plants in shade, depriving them of sunlight.
2. Plants and algae are consumed by bacteria in the water.
3. Underwater plants die.
4. Nutrients accumulate in the water in excess due to runoff.
a. 3 - 4 - 1 - 2
b. 1 - 4 - 3 - 2
c. 4 - 1 - 3 - 2
d. 1 - 4 - 2 - 3
2. Which of the following is correctly stated?
a. Cultural eutrophication can lead to hypoxia in aquatic systems.
b. Hypoxia is the most common cause of eutrophication, but not cultural eutrophication.
c. Hypoxia is a condition in which aquatic organisms produce too much oxygen.
d. Hypoxia causes low dissolved oxygen concentrations to occur.
3. An ultraoligotrophic lake has been isolated and unchanged for more than a thousand years.
Since we have the ability to grow algae and turn it into oil that we can use to make gasoline, one
day a company buys this property and tries to grow algae in the lake. Given the lack of nutrients,
the algae just won't grow and the company goes out of business. Which statement below about
this lake is correct?
a. The lake has high water quality.
b. The lake has undergone eutrophication
c. This question can't be answered without measuring the concentration of specific nutrients.
d. The lake has low water quality.
Kindly answer the 3 questions. I'll give you an upvote please.
.
Structure Hydra has a tubular body, with only one opening the mouth.pdf
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g
Solution
i) 10-2= centi, so 6.35cL
ii)10-6=micro, so 6.5s
iii) 10-6=micro, so 95m
iv) 10-9=nano, so 4.23nm
v) 103=kilo, so 12.5*10-8*103g=12.5*10-5g, 10-6=micro, so 125g.
Quality management ensures that an organization, product or service .pdf
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M
Solution
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M.
public class Deque {private class Node {public int data;public.pdf
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..
Microsoft Excel is a spreadsheet program used to store and retrieve .pdf
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent.
Solution
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent..
Mean = xffS = Sqrt((x-x)2fn)intervalMidpoint(x)frequency.pdf
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3
Solution
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3.
Moles of Be = mass of Bemolar mass of Be= (0.33 g)(9.012 gmol).pdf
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further.
Solution
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further..
Lyme disease is a bacterial infection which is caused by bacteria ca.pdf
Customers table
CREATE TABLE CUSTOMERS(ID INT PRIMARY KEY,LASTNAME
CHAR(10),FIRSTNAME CHAR(10),ADDRESS VARCHAR(30),CURRENTBALANCE
INT,CREDIT_LIMIT INT,SALESREP_ID INT);
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'124\',\'ADAMS\',\'SALLY\',\'481 OAK LANSING MI
49224\',\'818.75\',\'1000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'256\',\'SAMUELS\',\'ANN\',\'215 PETE GRANT MI
49219\',\'21.5\',\'1500\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'311\',\'CHARLES\',\'DON\',\'48 COLLEGE IRA MI
49034\',\'825.75\',\'1000\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'315\',\'DANIELS\',\'TOM\',\'914 CHERRY KENT MI
48391\',\'770.75\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'405\',\'WILLIAMS\',\'AL\',\'519 WATSON GRANT MI
49219\',\'402.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'412\',\'ADAMS\',\'SALLY\',\'16 ELM LANSING MI
49224\',\'1817.5\',\'2000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'522\',\'NELSON\',\'MARY,\'108 PINE ADA MI
49441\',\'98.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'567\',\'DINH\',\'TRAN\',\'808 RIDGE HARPER MI
48421\',\'402.4\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'587\',\'GALVEZ\',\'MARA\',\'512 PINE ADA MI
49441\',\'114.6\',\'1000\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'622\',\'MARTIN\',\'DAN\',\' 419 CHIP GRANT MI
49219\',\'1045.75\',\'1000\',\'3\');
Salesreps table
CREATE TABLE SALESREPS(ID INT,LASTNAME CHAR(10),FIRSTNAME
CHAR(10),ADDRESS VARCHAR(20),TOTAL _COMMISSION
INT(10),COMMISSION_RATE INT(10));
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'3\',\'JONES\',\'MARY\',\'123 MAIN GRANT MI 49219\',\'2150\',\'.05\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'6\',\'SMITH\',\'WILLIAM\',\'102 RAYMOND ADA MI
49441\',\'4912.5\',\'.07\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'12\',\'DIAZ\',\'MIGUEL\',\'419 HARPER LANSING MI
49224\',\'2150\',\'.05\');
Orders table
CREATE TABLE ORDERS(ID INT FOREIGN KEY,ORDER_DATE DATE,CUSTOMER
INT(10),SHIPPING_DATE DATE);
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12489\',\'02-JUL-11\',\'124,\'22-JUL-11\');
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12491\',\'02-JUL-11\',\'311\',\'22-.
import java.util.Scanner; import java.util.Random; public clas.pdf
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg
Solution
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg.
above molecule has 12 signlas. .pdf
above molecule has 12 signlas.
Solution
above molecule has 12 signlas..
Hi,I have added a loop for adding values to list. Highlighted the .pdf
Answer(a):
We have r(t)=ln19ti+2j+1t2k ------(1)
Velocity=v(t)=r\'(t)
Now we need todo differentation with respect to( 1)
we get
r\'(t)=(1/19t)19i+2j+2tk
19 gets cancellled
=(1/t)i+2j+2tk
Therefore r\'(t)=(1/t)i+2j+2tk----(2)
Comparing the above equation with a= xi+yj+zk
|a|=sq rt(x2+y2+z2)
Therfore x=1/t y=2 z=2t
To find ||r\'(t)||=Sq rt[(1/t)2+(2)2+(2t)2
=sq rt[{1/t2+22+(2t)2
=sq rt[1/t2+4+4t2)
=sq rt[(1+2t2)2/t2
=eliminating sq root we have
=1+2t2/t
=1/t+2t
Therefore |r\'t|=1/t+2t
Answer(b)
Therefore speed of t=s(t)= |r\'t|
=1/t+2t
Answer(C):
To find c we need to integrate r\'t and then apply limits
Therefore integrating wrt t
we have
ln(t)i+2tj+t2k
Now apply limits
upper limit- lower limit to the integral evaluated above
={ln(3)i+2(3)j+(3)2k}-{ln(2)i+2(2)j+(2)2k}
=ln3i+6j+9k-ln2i-4j-4k
=ln(3/2)i+2j+5k
since ln(a)-ln(b)=lna/b
Answer:ln(3/2)i+2j+5k
Solution
Answer(a):
We have r(t)=ln19ti+2j+1t2k ------(1)
Velocity=v(t)=r\'(t)
Now we need todo differentation with respect to( 1)
we get
r\'(t)=(1/19t)19i+2j+2tk
19 gets cancellled
=(1/t)i+2j+2tk
Therefore r\'(t)=(1/t)i+2j+2tk----(2)
Comparing the above equation with a= xi+yj+zk
|a|=sq rt(x2+y2+z2)
Therfore x=1/t y=2 z=2t
To find ||r\'(t)||=Sq rt[(1/t)2+(2)2+(2t)2
=sq rt[{1/t2+22+(2t)2
=sq rt[1/t2+4+4t2)
=sq rt[(1+2t2)2/t2
=eliminating sq root we have
=1+2t2/t
=1/t+2t
Therefore |r\'t|=1/t+2t
Answer(b)
Therefore speed of t=s(t)= |r\'t|
=1/t+2t
Answer(C):
To find c we need to integrate r\'t and then apply limits
Therefore integrating wrt t
we have
ln(t)i+2tj+t2k
Now apply limits
upper limit- lower limit to the integral evaluated above
={ln(3)i+2(3)j+(3)2k}-{ln(2)i+2(2)j+(2)2k}
=ln3i+6j+9k-ln2i-4j-4k
=ln(3/2)i+2j+5k
since ln(a)-ln(b)=lna/b
Answer:ln(3/2)i+2j+5k.
D. ICANNThe Internet Assigned Numbers Authority (IANA) is a depart.pdf
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y
Solution
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y.
Cracking of petrolium In petroleum geology and chemistry, cracki.pdf
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9
Solution
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9.
can you provide any aditional information apart from thisSolutio.pdf
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male
Solution
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male.
brasstinSolutionbrasstin.pdf
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6
Solution
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6.
AnswerMajority of Americans have faster advancing into above 85 y.pdf
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i
Solution
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i.
AnswerA compilation error is generated The method f(int) is und.pdf
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-)
Solution
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-).
a) There is a cluster of SNP’s with similar p value in the intron .pdf
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s)
Solution
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s).
A HRIS, which is otherwise called a human resource information syste.pdf
/*
* Sorter class contains static methods for bubble sort,
* insertion sort,selection sort and quick sort.
*
* */
//Sorter.java
public class Sorter
{
//Bubble sort
public static int[] bubbleSort(int list[])
{
for (int c = 0; c < ( list.length - 1 ); c++)
{
for (int d = 0; d < list.length - c - 1; d++)
{
if (list[d]>list[d+1])
/* For ascending order*/
{
int temp= list[d];
list[d]=list[d+1];
list[d+1]=temp;
}
}
}
return list;
}
//Selection sort
public static int[] selectionSort(int[] list)
{
for (int i = 0; i < list.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < list.length; j++)
if (list[j] 0 ; j--)
{
if(list[j] pivot)
{
j--;
}
if (i <= j) {
exchangeNumbers(array,i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort()
if (lowerIndex < j)
quickSort(array,lowerIndex, j);
if (i < higherIndex)
quickSort(array,i, higherIndex);
}
//exchange elements at index i and j
private static void exchangeNumbers(int array[],int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
----------------------------------------------------------------------------------------------------------------
/**
* The java program Driver that calls the methods
* of Sorter class to sort 8 element array for bubble sort,
* insertion and selection sort.
* Call quicksort for 16 element array and print
* values before and after sorting .
* */
//Driver.java
public class Driver
{
public static void main(String[] args)
{
int arr1[]={1,7,3,2,0,5,0,8};
int arr2[]={1,7,3,2,0,5,0,8};
int arr3[]={1,7,3,2,0,5,0,8};
System.out.println(\"Before bubble sort\");
for (int i = 0; i < arr1.length; i++)
System.out.printf(\"%5d\",arr1[i]);
int bsort[]=Sorter.bubbleSort(arr1);
System.out.println(\"\ After bubble sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",bsort[i]);
System.out.println(\"\ Before insertion sort\");
for (int i = 0; i < arr2.length; i++)
System.out.printf(\"%5d\",arr2[i]);
int isort[]=Sorter.insertionSort(arr2);
System.out.println(\"\ After insertion sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",isort[i]);
System.out.println(\"\ Before selection sort\");
for (int i = 0; i < arr3.length; i++)
System.out.printf(\"%5d\",arr3[i]);
int ssort[]=Sorter.insertionSort(arr3);
System.out.println(\"\ After selection sort\");
for (int i = 0; i < ssort.length; i++)
System.out.printf(\"%5d\",ssort[i]);
int arr4[]={22,36,6,79,26,45,75,13,31,62,27,76,33,16,62,47};
System.out.println(\"\ Before quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
Sorter.quickSort(arr4, 0, arr4.length-1);
System.out.println(\"\ After quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
}
}
----------------------------------------------------------------------------------------------------------------
Sample output:
Before bubble sort
1 7 3 2 0 5 0 8
After bubble sort
0 0 1 2 3 5 7 8
Before insertion sort
1 7 3 2 0 5 0 8
After i.
4 (110000)^3 = 4 s^2 0.01s = 10^-5 MoleslitSolution.pdf
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles.
Solution
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles..
2009.88Solution2009.88.pdf
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms
Solution
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms.
It is not that simple because TM cmplxs often have two or three band.pdf
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles
Solution
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles.
A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance
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Quality management ensures that an organization, product or service .pdf
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M
Solution
Given:
15.0ml of concentrated (11.6M) HCl into a final volume of 500mL
First step you have to convert ml to l (litters)
15.0ml divided by 1000ml=0.015L
500ml divided by 1000ml=0.5L
Then, use this formula to solve concentration: M1V1=M2V2
(0.5L)M1=(0.015L)(11.6M)
M1=(0.015L)(11.6) divided by (0.5L)
M1=0.348M (use 3 sigfigs as a final answer)
Final answer is 0.348M.
public class Deque {private class Node {public int data;public.pdf
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval.
Solution
Given interval is `[0,2pi]` .
And given identity is `secx+tanx=1`
which we can write as `1/(cosx)+(sinx)/(cosx)=1`
`or` , `(1+sinx)/(cosx)=1`
or, `(1+sinx)=cosx`
or, `1+sinx-cosx=0`
or, `1+2sin(x/2)cos(x/2)-(1-2sin^2(x/2))=0`
or, `1+2sin(x/2)cos(x/2)-1+2sin^2(x/2)=0`
or, `2sin(x/2)cos(x/2)+2sin^2(x/2)=0`
or, `2sin(x/2)[cos(x/2)+sin(x/2)]=0`
`or,` `2sin(x/2)=0` and `cos(x/2)+sin(x/2)=0` .
Now, `2sin(x/2)=0` `rArrsin(x/2)=0` `rArrsin(x/2)=sin(npi)` , n=0,1,2. As per the need of our
problem.
`rArr x/2=npi` or, `x=2npi` for n=0,1 as our interval is `[0,2pi]` .
So, `x=0` or, `x=2pi` .
Now, if `cos(x/2)+sin(x/2)=0` `rArr1+(sin(x/2))/cos(x/2)=0`
`rArrtan(x/2)=-1` or, `tan(x/2)=tan(pi/2+pi/4)`
or, `x/2=(pi/2+pi/4)`
or, `x=pi+pi/2`
or, `x=3pi/2` .
Clearly, `x=0` and `x=2pi` are the solutions satisfying our given identity over the given interval..
Microsoft Excel is a spreadsheet program used to store and retrieve .pdf
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent.
Solution
Diffrence between totipotent and pluripotent cells are :
Totipotent cells can form all the cell types in a body, plus the extraembryonic, or placental, cells.
Embryonic cells within the first couple of cell divisions after fertilization are the only cells that
are totipotent. Pluripotent cells on the other hand can give rise to all of the cell types that make
up the body; embryonic stem cells are considered pluripotent..
Mean = xffS = Sqrt((x-x)2fn)intervalMidpoint(x)frequency.pdf
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3
Solution
Density = mass/volume
= 32 g /4 cm3 = 8 g/cm3.
Moles of Be = mass of Bemolar mass of Be= (0.33 g)(9.012 gmol).pdf
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further.
Solution
Do you want to evaluate 2sin 18 cos 18?
There are a lot of apporaches for this problem. Here I show you one approach:
Let x = 18 degrees.
sin 2x = sin 36 = cos 54 = cos 3x, using co-function property
Expand,
2sin x cos x = cos x cos 2x - sin x sin 2x
Cancel cos x and let u = sin x = sin 18,
2u = 1 - 2u^2 - 2u^2 = 0, sinc cos 2x = 1 - 2sin^2 x
Collect all terms in one side,
4u^2 + 2u - 1 = 0
Apply quadratic formula,
u = (1/8)[-2+sqrt(4+16)] = (1/4)(-1+sqrt(5))
cos 18 = sqrt[1-u^2] = (1/4)sqrt[10+2sqrt(5)]
2sin 18 cos 18 = 2(1/4)(-1+sqrt(5))(1/4)sqrt[10+2sqrt(5)] which can be simplified further..
Lyme disease is a bacterial infection which is caused by bacteria ca.pdf
Customers table
CREATE TABLE CUSTOMERS(ID INT PRIMARY KEY,LASTNAME
CHAR(10),FIRSTNAME CHAR(10),ADDRESS VARCHAR(30),CURRENTBALANCE
INT,CREDIT_LIMIT INT,SALESREP_ID INT);
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'124\',\'ADAMS\',\'SALLY\',\'481 OAK LANSING MI
49224\',\'818.75\',\'1000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'256\',\'SAMUELS\',\'ANN\',\'215 PETE GRANT MI
49219\',\'21.5\',\'1500\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'311\',\'CHARLES\',\'DON\',\'48 COLLEGE IRA MI
49034\',\'825.75\',\'1000\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'315\',\'DANIELS\',\'TOM\',\'914 CHERRY KENT MI
48391\',\'770.75\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'405\',\'WILLIAMS\',\'AL\',\'519 WATSON GRANT MI
49219\',\'402.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'412\',\'ADAMS\',\'SALLY\',\'16 ELM LANSING MI
49224\',\'1817.5\',\'2000\',\'3\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'522\',\'NELSON\',\'MARY,\'108 PINE ADA MI
49441\',\'98.75\',\'1500\',\'12\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'567\',\'DINH\',\'TRAN\',\'808 RIDGE HARPER MI
48421\',\'402.4\',\'750\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'587\',\'GALVEZ\',\'MARA\',\'512 PINE ADA MI
49441\',\'114.6\',\'1000\',\'6\');
INSERT INTO
CUSTOMERS(ID,LASTNAME,FIRSTNAME,ADDRESS,CURRENTBALANCE,CREDIT_LI
MIT,SALESREP_ID) VALUES(\'622\',\'MARTIN\',\'DAN\',\' 419 CHIP GRANT MI
49219\',\'1045.75\',\'1000\',\'3\');
Salesreps table
CREATE TABLE SALESREPS(ID INT,LASTNAME CHAR(10),FIRSTNAME
CHAR(10),ADDRESS VARCHAR(20),TOTAL _COMMISSION
INT(10),COMMISSION_RATE INT(10));
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'3\',\'JONES\',\'MARY\',\'123 MAIN GRANT MI 49219\',\'2150\',\'.05\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'6\',\'SMITH\',\'WILLIAM\',\'102 RAYMOND ADA MI
49441\',\'4912.5\',\'.07\');
INSERT INTO
SALESREPS(ID,LASTNAME,FIRSTNAME,ADDRESS,TOTAL_COMMISSION,COMMISSI
ON_RATE) VALUES(\'12\',\'DIAZ\',\'MIGUEL\',\'419 HARPER LANSING MI
49224\',\'2150\',\'.05\');
Orders table
CREATE TABLE ORDERS(ID INT FOREIGN KEY,ORDER_DATE DATE,CUSTOMER
INT(10),SHIPPING_DATE DATE);
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12489\',\'02-JUL-11\',\'124,\'22-JUL-11\');
INSERT INTO ORDERS(ID,ORDER_DATE,CUSTOMER,SHIPPING_DATE)
VALUES(\'12491\',\'02-JUL-11\',\'311\',\'22-.
import java.util.Scanner; import java.util.Random; public clas.pdf
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg
Solution
Assuming that the system is a closed system, the mass of the system is constant througout the
process.Hence the volume of the system doesnot change making it an isochoric process.
So, win = 0 kJ/kg.
above molecule has 12 signlas. .pdf
above molecule has 12 signlas.
Solution
above molecule has 12 signlas..
Hi,I have added a loop for adding values to list. Highlighted the .pdf
Answer(a):
We have r(t)=ln19ti+2j+1t2k ------(1)
Velocity=v(t)=r\'(t)
Now we need todo differentation with respect to( 1)
we get
r\'(t)=(1/19t)19i+2j+2tk
19 gets cancellled
=(1/t)i+2j+2tk
Therefore r\'(t)=(1/t)i+2j+2tk----(2)
Comparing the above equation with a= xi+yj+zk
|a|=sq rt(x2+y2+z2)
Therfore x=1/t y=2 z=2t
To find ||r\'(t)||=Sq rt[(1/t)2+(2)2+(2t)2
=sq rt[{1/t2+22+(2t)2
=sq rt[1/t2+4+4t2)
=sq rt[(1+2t2)2/t2
=eliminating sq root we have
=1+2t2/t
=1/t+2t
Therefore |r\'t|=1/t+2t
Answer(b)
Therefore speed of t=s(t)= |r\'t|
=1/t+2t
Answer(C):
To find c we need to integrate r\'t and then apply limits
Therefore integrating wrt t
we have
ln(t)i+2tj+t2k
Now apply limits
upper limit- lower limit to the integral evaluated above
={ln(3)i+2(3)j+(3)2k}-{ln(2)i+2(2)j+(2)2k}
=ln3i+6j+9k-ln2i-4j-4k
=ln(3/2)i+2j+5k
since ln(a)-ln(b)=lna/b
Answer:ln(3/2)i+2j+5k
Solution
Answer(a):
We have r(t)=ln19ti+2j+1t2k ------(1)
Velocity=v(t)=r\'(t)
Now we need todo differentation with respect to( 1)
we get
r\'(t)=(1/19t)19i+2j+2tk
19 gets cancellled
=(1/t)i+2j+2tk
Therefore r\'(t)=(1/t)i+2j+2tk----(2)
Comparing the above equation with a= xi+yj+zk
|a|=sq rt(x2+y2+z2)
Therfore x=1/t y=2 z=2t
To find ||r\'(t)||=Sq rt[(1/t)2+(2)2+(2t)2
=sq rt[{1/t2+22+(2t)2
=sq rt[1/t2+4+4t2)
=sq rt[(1+2t2)2/t2
=eliminating sq root we have
=1+2t2/t
=1/t+2t
Therefore |r\'t|=1/t+2t
Answer(b)
Therefore speed of t=s(t)= |r\'t|
=1/t+2t
Answer(C):
To find c we need to integrate r\'t and then apply limits
Therefore integrating wrt t
we have
ln(t)i+2tj+t2k
Now apply limits
upper limit- lower limit to the integral evaluated above
={ln(3)i+2(3)j+(3)2k}-{ln(2)i+2(2)j+(2)2k}
=ln3i+6j+9k-ln2i-4j-4k
=ln(3/2)i+2j+5k
since ln(a)-ln(b)=lna/b
Answer:ln(3/2)i+2j+5k.
D. ICANNThe Internet Assigned Numbers Authority (IANA) is a depart.pdf
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y
Solution
6x3(3x2+5y) = 6x3(3x2) + 6x3(5y) = 18x5 + 30x3y.
Cracking of petrolium In petroleum geology and chemistry, cracki.pdf
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9
Solution
4s^3 = 4 * (1.17*10^-3)^3 = 6.41*10^-9.
can you provide any aditional information apart from thisSolutio.pdf
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male
Solution
2.no staining, positive staining
3.stain positively, no staining
4. stain positively for female embryo, no staining for male embryo.
XX : XX female ; XY : XO male.
brasstinSolutionbrasstin.pdf
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6
Solution
11)
(3n^2 - n) / n^2
_________________
(n^2 -1 )/( n +1)
=
n*(3n -1) / n^2
_________________
(n+1)(n -1 )/( n +1)
=
(3n -1) / n
_________________
(n -1 )
=
(3n -1)
_________________
(n^2 -n )
12. The width w of a rectangular swimming pool is 7x2. The area A of the pool is 7x3 – 42x2.
What is an expression for the length of the pool? Show your work. Determine whether each
question is biased. Explain your answer.
length *width =area
length *7x^2 =7x^3 -42x^2
length =(7x^3 -42x^2)/7x^2
length =x -6.
AnswerMajority of Americans have faster advancing into above 85 y.pdf
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i
Solution
1) Vector\'s components in x - direction : 75*cos (35) = 61.436
Vector\'s components in y - direction : 75*sin (35) = 43.0182
2)
Vector is ( 2 -i) - (-3 + 8i) = 5 -9 i
so magnitude of the horizontal component = 5
3)
z = 5 + 12i
magnitude of vector = sqrt( 5^2 + 12^2) = 13
direction = atan ( 12/5) =67.3801 degrees
resultant vector\'s magnitude is 13 and it\'s direction is 67.3801 degrees.
13 <67.3801 is equivalent to z = 5 + 12i.
AnswerA compilation error is generated The method f(int) is und.pdf
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-)
Solution
1) Convert angstroms to cm:
4.09 angstroms x 10-8 cm/1 angstrom = 4.09 x 10-8 cm
2) Determine the volume of the unit cube:
(4.09 x 10-8 cm)^3 = 6.842 10-23 cm3
3) Determine the mass of the metal in the unit cube:
10.5 g/cm3 times 6.842 x 10-23 cm3 = 7.184 x 10-22 g
4) Determine atomic weight (based on 4 atoms per unit cell):
7.184 x 10-22 g is to 4 atoms as x grams is to 6.022 x 1023 atoms. To find x, you just have to
do cross multiplication.
x = 108.16 g/mol, so the element is Ag
Hope this helps! :-).
a) There is a cluster of SNP’s with similar p value in the intron .pdf
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s)
Solution
K1+ + CrO42- + Ag1+NO31-
Now lets go through the solubility rules and cross out everythingsoluble:
K1+ + CrO42- +Ag1+NO31-
Group 1A metals are always soluble (Rule 1)
Compounds of NO31- are always soluble (Rule2)
So we are left with Ag1+ andCrO42- to react with each other asfollows:
2Ag1+(aq) +CrO42-(aq) ----->Ag2CrO4(s).
A HRIS, which is otherwise called a human resource information syste.pdf
/*
* Sorter class contains static methods for bubble sort,
* insertion sort,selection sort and quick sort.
*
* */
//Sorter.java
public class Sorter
{
//Bubble sort
public static int[] bubbleSort(int list[])
{
for (int c = 0; c < ( list.length - 1 ); c++)
{
for (int d = 0; d < list.length - c - 1; d++)
{
if (list[d]>list[d+1])
/* For ascending order*/
{
int temp= list[d];
list[d]=list[d+1];
list[d+1]=temp;
}
}
}
return list;
}
//Selection sort
public static int[] selectionSort(int[] list)
{
for (int i = 0; i < list.length - 1; i++)
{
int index = i;
for (int j = i + 1; j < list.length; j++)
if (list[j] 0 ; j--)
{
if(list[j] pivot)
{
j--;
}
if (i <= j) {
exchangeNumbers(array,i, j);
//move index to next position on both sides
i++;
j--;
}
}
// call quickSort()
if (lowerIndex < j)
quickSort(array,lowerIndex, j);
if (i < higherIndex)
quickSort(array,i, higherIndex);
}
//exchange elements at index i and j
private static void exchangeNumbers(int array[],int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
}
----------------------------------------------------------------------------------------------------------------
/**
* The java program Driver that calls the methods
* of Sorter class to sort 8 element array for bubble sort,
* insertion and selection sort.
* Call quicksort for 16 element array and print
* values before and after sorting .
* */
//Driver.java
public class Driver
{
public static void main(String[] args)
{
int arr1[]={1,7,3,2,0,5,0,8};
int arr2[]={1,7,3,2,0,5,0,8};
int arr3[]={1,7,3,2,0,5,0,8};
System.out.println(\"Before bubble sort\");
for (int i = 0; i < arr1.length; i++)
System.out.printf(\"%5d\",arr1[i]);
int bsort[]=Sorter.bubbleSort(arr1);
System.out.println(\"\ After bubble sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",bsort[i]);
System.out.println(\"\ Before insertion sort\");
for (int i = 0; i < arr2.length; i++)
System.out.printf(\"%5d\",arr2[i]);
int isort[]=Sorter.insertionSort(arr2);
System.out.println(\"\ After insertion sort\");
for (int i = 0; i < bsort.length; i++)
System.out.printf(\"%5d\",isort[i]);
System.out.println(\"\ Before selection sort\");
for (int i = 0; i < arr3.length; i++)
System.out.printf(\"%5d\",arr3[i]);
int ssort[]=Sorter.insertionSort(arr3);
System.out.println(\"\ After selection sort\");
for (int i = 0; i < ssort.length; i++)
System.out.printf(\"%5d\",ssort[i]);
int arr4[]={22,36,6,79,26,45,75,13,31,62,27,76,33,16,62,47};
System.out.println(\"\ Before quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
Sorter.quickSort(arr4, 0, arr4.length-1);
System.out.println(\"\ After quicksort sort\");
for (int i = 0; i < arr4.length; i++)
System.out.printf(\"%5d\",arr4[i]);
}
}
----------------------------------------------------------------------------------------------------------------
Sample output:
Before bubble sort
1 7 3 2 0 5 0 8
After bubble sort
0 0 1 2 3 5 7 8
Before insertion sort
1 7 3 2 0 5 0 8
After i.
4 (110000)^3 = 4 s^2 0.01s = 10^-5 MoleslitSolution.pdf
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles.
Solution
Xe (I think) because it is a bigger molecule the electron cloud has transient dipoles..
2009.88Solution2009.88.pdf
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms
Solution
we have, P4 + 10Cl2 -------> 5PCl5 10 moles of CL2 gives 5 moles of PCl5 now,
52/71 moles will give = 52/71 x 5/10 = 0.366 moles = 76.352 gms.
It is not that simple because TM cmplxs often have two or three band.pdf
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles
Solution
Let moles of SbCl5 be x 1/Kb=[SbCl5]/[cl2][SbCl3] 1/2.5*10^-
2=(x/2.5)/((0.15/2.5)*(0.27/2.5)) x=0.648 moles.
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- 1. 1. It will be absorbed by plant. Or remain in soil that can be absorbed by plant later. Solid fertilizers is mainly composed of Urea, when it is added with little amount of water to soil, it get hydrolyzed and converted to ammonia and Co2. This process is normally take 72 hours. Basing on water content and extra water might cause urea to run off, leading to loss of urea. CO(NH2)2 + H2O + urease 2NH3 +CO2 (urea) 2. When added more water to the soil, this water will resist the movement of microorganisms. This is because oxygen is resisted to move freely, due to which microbial activity is restricted as well. Lack of oxygen will cause to microorganisms metabolism is been effected. Normal pathways are disrupted due to lack of O2. Solution 1. It will be absorbed by plant. Or remain in soil that can be absorbed by plant later. Solid fertilizers is mainly composed of Urea, when it is added with little amount of water to soil, it get hydrolyzed and converted to ammonia and Co2. This process is normally take 72 hours. Basing on water content and extra water might cause urea to run off, leading to loss of urea. CO(NH2)2 + H2O + urease 2NH3 +CO2 (urea) 2. When added more water to the soil, this water will resist the movement of microorganisms. This is because oxygen is resisted to move freely, due to which microbial activity is restricted as well. Lack of oxygen will cause to microorganisms metabolism is been effected. Normal pathways are disrupted due to lack of O2.