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3 years ago

What is the direct variation equation if f(x) varies directly with x and f(x)=−18 when x = 3?

Mathematics

1 answer:

kondor19780726 [428]3 years ago

8 0

Answer:

your answer is B becausewe can find k when given any point by dividing the y-coordinate by the x-coordinate. For example, if y varies directly as x, and y = 6 when x = 2, the constant of variation is k = - 3.

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If F(x)=2x/7+4, which of the following of the inverse of F(x)?

lions [1.4K]

Hello,

Answer D

y=2x/7+4
x=2y/7+4
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3 years ago

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IRISSAK [1]

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Find the limit, if it exists. (If an answer does not exist, enter DNE.)

gavmur [86]

Answer:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)=14$

Step-by-step explanation:

to find the limit:

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\right)$

we need to first rationalize our expression.

$\dfrac{x^2+y^2}{\sqrt{x^2+y^2+49}-7}\left(\dfrac{\sqrt{x^2+y^2+49}+7}{\sqrt{x^2+y^2+49}+7}\right)$

$\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(\sqrt{x^2+y^2+49}\,)^2-7^2}$

$\dfrac{(x^2+y^2)(\sqrt{x^2+y^2+49}+7)}{(x^2+y^2)}$

$\sqrt{x^2+y^2+49}+7$

Now this is our simplified expression, we can use our limit now.

$\lim\limits_{(x,y)\rightarrow(0,0)}\left(\sqrt{x^2+y^2+49}+7\right)\\\sqrt{0^2+0^2+49+7}\\7+7\\14$

Limit exists and it is 14 at (0,0)

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3 years ago