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3 years ago

What is equivalent to 5 to the 2nd power times 5 to the 2nd power

Mathematics

2 answers:

Semenov [28]3 years ago

5 0

Answer:

625

Step-by-step explanation:

adelina 88 [10]3 years ago

3 0

<em>5 to the 2nd power times 5 to the 2nd power is equivalent to 625</em>

<h2>Explanation:</h2>

In this case, we have the following statement:

<em>5 to the 2nd power times 5 to the 2nd power. </em>In a mathematical language:

$5^2\times 5^2$

From product rules, we know that:

$a^n \cdot a^m=a^{n+m}$

Applying this rule to our problem:

$5^2\times 5^2=5^{2+2}=5^4=625$

<em>5 to the 2nd power times 5 to the 2nd power is equivalent to 625</em>

<h2>Learn more:</h2>

Rules of exponents: brainly.com/question/12140557

#LearnWithBrainly

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3 years ago

Find the product of 3 1/8 and 1 2/7. Express your answer in simplest form

earnstyle [38]

Answer:

Step-by-step explanation:

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3 years ago

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Westkost [7]

Find the center of the door.
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Find the center of the towel rack
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Note: I used decimals, this can also be done with fractions.

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3 years ago

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Find the absolute extreme of the function on the closed interval. y=9e^xsinx , [0,pi]

Molodets [167]

For convenience sake, I will let $y=f(x)=9e^{x}\sin x$

First, we evaluate the function at the endpoints of the interval.

$f(0)=9e^{0}\sin(0)=0\\\\f(\pi)=9e^{\pi}\sin(\pi)=0$

Then, we need to find the critical points.

We can start by taking the derivative using the power rule.

$f'(x)=9e^{x} \frac{d}{dx} \sin x+9\sin x \frac{d}{dx} e^{x}=9e^{x}(\cos x+\sin x)$

Setting this equal to 0,

$9e^x (\cos x+\sin x)=0$

Since $9e^x > 0$ , we can divide both sides by $9e^x$ .

$\cos x+\sin x=0\\\\\sin x=-\cos x\\\\\tan x=-1\\\\x=\frac{3\pi}{4}$

$f\left(\frac{3\pi}{4} \right)=9e^{3\pi/4}\sin \left(\frac{3\pi}{4} \right)=\frac{9\sqrt{2}e^{3\pi/4}}{2}$

So, the absolute minimum is $\boxed{\left(\frac{3\pi}{4}, \frac{9\sqrt{2}e^{3\pi/4}}{\sqrt{2}} \right)}$ and the absolute minima are $\boxed{(0,0), (\pi, 0)}$

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2 years ago