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The document provides solutions to 7 cryptarithm problems. For each problem, it states the values that letters A and B can represent, then uses logical deductions based on place values to determine the values of A and B that satisfy the cryptarithm equation. The last problem shows that cryptarithm 4 x AB = CAB has no solution because there is no digit whose units place is equal to itself multiplied by 4.
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an equation of the form
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Quadratic equations
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ax2 + bx + c = 0
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Complete Quadratic
3x2 + 5x + 6 = 0
Incomplete/Pure Quadratic Equation
3x2 - 6 = 0
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x2 – 4 = 0
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5x² - 11 = 49
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x² = 12
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Place all terms in the left member of the equation, so that the right member is zero.
Factor the left member.
Set each factor that contains the unknown equal to zero.
Solve each of the simple equations thus formed.
Check the answers by substituting them in the original equation.
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Solution:
x² = 6x – 8
x² - 6x + 8 = 0
(x – 4)(x – 2) = 0
x – 4 = 0 | x – 2 = 0
x = 4 x = 2
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Write the equation with the variable terms in the left member and the constant term in the right member.
If the coefficient of x² is not 1, divide every term by this coefficient so as to make the coefficient of x² equal to 1.
Take one-half the coefficient of x, square this quantity, and add the result to both members.
Find the square root of both members, placing a ± sign before the square root of the right member.
Solve the resulting equation for x.
10.Example: x² - 8x + 7 = 0
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Example: 3x² - 2x - 7 = 0
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Bilangan Bulat Matematika Kelas 7
Materi Bilangan Bulat Matematika Kelas 7
Terdiri dari :
Penjumlahan Bilangan Bulat
Pengurangan Bilangan Bulat
Perkalian Bilangan Bulat
Pembagian Bilangan Bulat
Pecahan Bilangan Bulat
Desimal
KPK dan FPB
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Who wants to eat 10 toffees
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C26
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The document defines various terms related to quadrilaterals and regular polygons. It then provides solutions to 19 questions involving calculating missing angle measures, identifying properties, and determining the number of sides of regular polygons given the measure of each interior angle. The questions cover properties of quadrilaterals like total angle sum, relationships between adjacent/opposite angles and sides, using angle measures to find unknown angles, and properties of regular polygons.
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C5 5.1
- 1. Solve each of the following cryptarithms. Question 1. Solution: Values of A and B be from 0 to 9 In ten’s digit 3 + A = 9 ∴ A = 6 or less. ∴ 7 + B = A = 6 or less ∴ 7 + 9 or 8 = 16 or 15 ∴ But it is two digit number B = 8 Then A = 5 Question 2. Solution: Values of A and B can be between 0 and 9 In tens digit, A + 3 = 9 ∴ A = 9 – 3 = 6 or less than 6 In ones unit B + 7 = A = 6or less ∴ 7 + 9 or 8 = 16 or 15 But it is two digit number ∴ B = 8 and ∴ A = 5 Question 3. Solution: Value of A and B can be between 0 and 9 In units place.
- 2. 1+B = 0 ⇒1+B = 10 ∴ B = 10 – 1 = 9 and in tens place 1 + A + 1 = B ⇒ A + 2 = 9 ⇒ A = 9 – 2 = 7 Question 4. Solution: Values of A and.B can be between 0 and 9 In units place, B+1 = 8 ⇒ B = 8-1=7 In tens place A + B= 1 or A + B = 11 ⇒ A + 7 = 11 ⇒ A =11-7 = 4 Question 5. Solution: Values of A and B can be between 0 and 9 In tens place, 2 + A = 0 or 2 + A=10 A = 10-2 = 8 In units place, A + B = 9 ⇒ 8 + B = 9 ⇒ B = 9- 8 = 1 Question 6. Solution: Values of A and B can be between 0 and 9
- 3. In hundreds place, Question 7. Show that cryptarithm 4 x (overline { AB } =overline { CAB }) does not have any solution. Solution: It means that 4 x B is a numebr whose units digit is B Clearly, there is no such digit Hence the given cryptarithm has no solution.