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alisha [4.7K]
1 year ago
12

Which equation is the inverse of 5 y + 4 = (x + 3) squared + one-half?

Mathematics
1 answer:
8_murik_8 [283]1 year ago
4 0

The inverse of the given function is y = √5x + 7/2 - 3

<h3>Inverse of a function</h3>

Given the function expressed below;

5y + 4 = (x+3)² +1/2

5y = (x+3)² - 7/2

y = 1/5(x+3)² - 7/10

Replace y as x

x = 1/5(y+3)² - 7/10

Make y the subject of the formula

5x = (y+3)²- 7/2

(y+3)² = 5x + 7/2

y+3 = √5x + 7/2

y = √5x + 7/2 - 3

Hence the inverse of the given function is y = √5x + 7/2 - 3

learn more on inverse of a function here: brainly.com/question/3831584

#SPJ1

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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=
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Answer:

The function \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}} is continuous at x = 36.

Step-by-step explanation:

We need to follow the following steps:

The function is:

\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}

The function is continuous at point x=36 if:

  1. The function \\ f(x) exists at x=36.
  2. The limit on both sides of 36 exists.
  3. The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}

\\ f(36) = \frac{36*6}{900} = \frac{6}{25}

The limit of the \\ f(x) is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}

\\ \lim_{x \to 36} f(x) = \frac{6}{25}

Since

\\ f(36) = \frac{6}{25}

And

\\ \lim_{x \to 36} f(x) = \frac{6}{25}

Then, the function \\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}} is continuous at x = 36.

8 0
3 years ago
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Answer:

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Step-by-step explanation:

-48 + 37 - 28 - 1

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-40

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