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kennyjortiz

kennyjortiz

09.12.2020 •

Mathematics

Please help me with 4/a, b, c and 5/b

Please help me with 4/a, b, c and 5/b

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Ответ:

bransom4206

12.02.2022

Answers:4a.) 49x^2-36y^2

49x^2-36y^2

4b.)

9a^2-25b^2

4c.) $\frac{1}{25}x^2-49$ 5b.) 4y^2-4xy-5x^2

4y^2-4xy-5x^2

Solution Steps:

__________________________________

4a.) $\bold{(7x-6y)(7x+6y)}$ :

- Multiplication can be transformed into difference of squares using the rule: (a-b)(a+b)=a^2-b^2 .

- So change the equation using the rule:

(7x)^2-(6y)^2

- Expand (7x)^2 and (6y)^2 :

(7x)^2=7^2x^2

(6y)^2=6^2y^2

- Calculate 7^2 and 6^2 :

7^2=49

6^2=36

So the end equation would be: 49x^2-36y^2 .

4b.) $\bold{(3a+5b)(3a-5b)}$ :

- Multiplication can be transformed into difference of squares using the rule: (a-b)(a+b)=a^2-b^2 .

- So change the equation using the rule:

(3a)^2-(5b)^2

- Expand (3a)^2 and (5b)^2 :

(3a)^2=3^2a^2

(5b)^2=5^2b^2

- Calculate 3^2 and 5^2 :

3^2=9

5^2=25

So the end equation would be: 9a^2-25b^2 .

4c.) $\bold{(\frac{1}{5}x-7)}$ × $\bold{(\frac{1}{5}+7)}$ :

- Multiplication can be transformed into difference of squares using the rule: (a-b)(a+b)=a^2-b^2 .

- So change the equation using the rule:

$(\frac{1}{5}x)^2-7^2$

- Expand $(\frac{1}{5}x)^2$ :

$(\frac{1}{5})^2x^2$

- Calculate $(\frac{1}{5})^2$ and 7^2 :

$(\frac{1}{5})^2=\frac{1}{25}$ 7^2=49

7^2=49

So the end equation would be: $\frac{1}{25}x^2-49$ .

5b.) $\bold{(4x-y)(4x+y)+(2x-y)^2-(5x+2y)(5x-2y)}$ :

- Consider (4x-y)(4x+y) . Multiplication can be transformed into difference of squares using the rule: (a-b)(a+b)=a^2-b^2 .

- So change the equation using the rule:

(4x)^2-y^2

- Expand (4x)^2 :

(4x)^2=4^2x^2

- Calculate 4^2 :

4^2=16

- Use binomial theorem (a-b)^2=a^2-2ab+b^2 to expand (2x-y)^2 :

4x^2-4xy+y^2

- Combine -y^2 and y^2 :

-y^2+y^2=0

- Consider (5x+2y)(5x-2y) . Multiplication can be transformed into difference of squares using the rule: (a-b)(a+b)=a^2-b^2 .

So change the equation using the rule:

(5x)^2-(2y)^2

- Expand (5x)^2 and (2y)^2 :

(5x)^2=5^2x^2

(2y)^2=2^2y^2

- Calculate 5^2 and 2^2 :

5^2=25

2^2=4

- Combine 20x^2 and -25x^2 :

20x^2+(-25x^2)=-5x^2

So the end equation would be: 4y^2-4xy-5x^2 .

__________________________________

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Ответ:

DarcieMATHlin6217

25.05.2020

60

Step-by-step explanation:

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