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Anastasy [175]
3 years ago
13

Some equations have the form y=kx, where k is a number.

Mathematics
2 answers:
lawyer [7]3 years ago
5 0

Answer:

C. y = x - 7

Step-by-step explanation:

Cloud [144]3 years ago
4 0

Hello from MrBillDoesMath!

Answer:

y = x - 7

Discussion:

Let's step through the answers one-by-one:

y/6 = x             => multiply both sides by 6

y = 6x

y = kx where k = 6

y/x = 1/2          => multiply both sides by x

y = (1/2) x

y = kx where k = 1/2

y = -0.5x

y = kx where k = -0.5

BUT y = x -7 cannot be written simply as y = kx because of the -7 term.

Thank you,

MrB

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Step-by-step explanation:

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\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}1+21+2+31+3+41+……………+8+91

Rationalizing the denominator, we get

\Rightarrow\left(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}+\sqrt{4}} \times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\right)+\cdots \ldots+\left(\frac{1}{\sqrt{8}+\sqrt{9}} \times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}\right)⇒(1+21×1−21−2)+(2+31×2−32−3)+(3+41×3−43−4)+⋯…+(8+91×8−98−9)

We know that,

\left(a^{2}-b^{2}\right)=(a+b)(a-b)(a2−b2)=(a+b)(a−b)

Now, on substituting the formula, we get,

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\Rightarrow \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{8}+\sqrt{9}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})⇒1+21+

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