If d = 20 and C = 20π

1 answer:

8 0

If d = 20 and C = 20π, then r = 10.
If d = 26 and C = 26π, then r = 13.
If r = 1/2 and C = π, then d = 1.
If r = 7 and C = 14π, then d = 14.
If r = 6 and C = 12π, then d = 12.
If r = 50 and C = 100π, then d = 100.
*Note, r= 1/2×d or d=2×r

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Which will result in a difference of squares? (–7x + 4)(–7x + 4) (–7x + 4)(4 – 7x) (–7x + 4)(–7x – 4) (–7x + 4)(7x – 4)

maw [93]

You can simply expand each product and see whether it gives you a difference of squares.

• $\mathsf{(-7x+4)\cdot (-7x+4)}$

That's actually $\mathsf{(-7x+4)^2:}$

$\mathsf{(-7x+4)^2}\\\\ \mathsf{=(-7x+4)\cdot (-7x+4)}\\\\ \mathsf{=(-7x+4)\cdot (-7x)+(-7x+4)\cdot 4}\\\\ \mathsf{=49x^2-28x-28x+16}$

$\mathsf{=49x^2-56x+16}$ ✖

which is not a difference of squares.

—————

• $\mathsf{(-7x+4)\cdot (4-7x)}$

$\mathsf{=(-7x+4)\cdot 4-(-7x+4)\cdot 7x}\\\\ \mathsf{=-28x+16-(-49x^2+28x)}\\\\ \mathsf{=-28x+16+49x^2-28x}$

$\mathsf{=49x^2-56x+16}$ ✖

which is not a difference of squares.

—————

• $\mathsf{(-7x+4)\cdot (-7x-4)}$

$\mathsf{=(-7x+4)\cdot (-7x)-(-7x+4)\cdot 4}\\\\ \mathsf{=49x^2-28x-(-28x+16)}\\\\ \mathsf{=49x^2-\diagup\!\!\!\!\! 28x+\diagup\!\!\!\!\! 28x-16}\\\\ \mathsf{=49x^2-16}$

$\mathsf{=(7x)^2-4^2}$ ✔

That is a difference of two squares.

—————

• $\mathsf{(-7x+4)\cdot (7x-4)}$

$\mathsf{=(-7x+4)\cdot 7x-(-7x+4)\cdot 4)}\\\\ \mathsf{=-49x^2+28x-(-28x+16)}\\\\ \mathsf{=-49x^2+28x+28x-16}$

$\mathsf{=-49x^2+56x-16}$ ✖

which is not a difference of squares.

—————

Only the third option will result in a difference of squares.

Answer: (− 7x + 4) · (− 7x − 4).

I hope this helps. =)

8 0

3 years ago

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xxTIMURxx [149]

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3 years ago

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3 years ago

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