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1 year ago

7

Help pls!!

1 answer:

nexus9112 [7]1 year ago

7 0

Answer: Choice D

Explanation:

The range is the set of all possible y outputs of a function. The highest y can go is y = 1, which occurs at the vertex. We can have y = 1 or y be smaller than this. Therefore, the range is $y \le 1$

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2 years ago

Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.

Sladkaya [172]

Answer:

The solution is $\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

Step-by-step explanation:

From the question

The function given is $f(x) = \frac{e^{2x}}{ 25 + e^{4x}} dx$

The indefinite integral is mathematically represented as

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx$

Now let $e^{2x} = u$

=> $\frac{du}{dx} 2e^{2x}$

=> $2 e^{2x}dx = du$

So

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx = \int\limits {\frac{1}{ 2(25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 5^2 + u^2)} } \, du$

$= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5} + C$

Now substituting for u

$\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

3 0

2 years ago

Please help mee pleaseee

nalin [4]

Answer:

The exact answer in terms of radicals is $x = 5*\sqrt[3]{25}$

The approximate answer is $x \approx 14.62009$ (accurate to 5 decimal places)

===============================================

Work Shown:

Let $y = \sqrt[5]{x^3}$

So the equation reduces to -7 = 8-3y

Let's solve for y

-7 = 8-3y

8-3y = -7

-3y = -7-8 ... subtract 8 from both sides

-3y = -15

y = -15/(-3) ... divide both sides by -3

y = 5

-----------

Since $y = \sqrt[5]{x^3}$ and y = 5, this means we can equate the two expressions and solve for x

$y = 5$

$\sqrt[5]{x^3} = 5$

$x^3 = 5^5$ Raise both sides to the 5th power

$x^3 = 3125$

$x = \sqrt[3]{3125}$ Apply cube root to both sides

$x = \sqrt[3]{125*25}$

$x = \sqrt[3]{125}*\sqrt[3]{25}$

$x = \sqrt[3]{5^3}*\sqrt[3]{25}$

$x = 5*\sqrt[3]{25}$

$x \approx 14.62009$

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3 years ago

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