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3 years ago

8

he sides of a square are five to the power of two fifths inches long. What is the area of the square?

2 answers:

Tresset [83]3 years ago

6 0

It would be <span>1.9 in.*1.9 in.
1.9*1.9</span>≈3.6 inches^2

lorasvet [3.4K]3 years ago

4 0

So, the area of a square is Length times width, but since it is a square, the value for the length is the wame as the widtch. So, since one side is 5^2/5, would you multiply <span>5^2/5 by itself. When you multiple something with the same bade, you would add the power. So, the answer would be

5^4/5
</span>

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-6 + u = -7 solve for the letter U

Illusion [34]

Answer:

U = -1

Step-by-step explanation:

-6 - 1 = -7

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3 years ago

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What is the inverse function of this function<br><br> f(x)=4x

Greeley [361]

F-1(x) = 4
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3 years ago

Find the zero(s) of each function algebraically.<br>f(x) = 8x-16<br>

vodomira [7]

the zero(s) of function $f(x)=8x-16$ is x=2

Step-by-step explanation:

We need to find the zero(s) of function algebraically.

We are given: $f(x)=8x-16$

To find the zeros we put the function equal to zero.

$8x-16=0\\Solving:\\8x=16\\x=\frac{16}{8}\\x=2$

So, the zero(s) of function $f(x)=8x-16$ is x=2

Keywords: zero(s) of function

Learn more about zero(s) of function at:

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3 years ago

What is the sum of this arithmetic series? <br><br> ASAP WILL MARK BRAINLIEST

Annette [7]

Answer:

ITS C OF THE ARITHMETIC SERIES !!

HOPE IT HELPS

Step-by-step explanation:

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3 years ago

Solve using a formula. Please don't guess, I would like professionals to take a look

Marianna [84]

<u>Let's take this problem step by step</u>:

<u>What we know:</u>

$x+y=4\\xy=-2$

<u>Before we solve, let's do one thing that will help us out greatly later down the road</u>:

$x+y=4\\(x+y)^2=4^2\\x^2+2xy+y^2=16\\x^2+2(xy)+y^2=16\\x^2+2(-2)+y^2=16\\x^2+4+y^2=16\\x^2+y^2=20$ <---<em> useful equation</em>

<u>Let's rearrange the problem a little bit:</u>

$x+\frac{x^3}{y^2}+\frac{y^3}{x^2} +y=\frac{x^3}{x^2} +\frac{x^3}{y^2}+\frac{y^3}{x^2}+\frac{y^3}{y^2}$

<u>Combine fractions of common denominators</u>:

$\frac{x^3+y^3}{x^2} +\frac{x^3+y^3}{y^2} =(x^3+y^3)*(\frac{1}{x^2}+\frac{1}{y^2} )$

<u>Now's let factor everything apart</u>:

$(x^3+y^3)=(x+y)(x^2-xy+y^2)\\\\\frac{1}{x^2}+\frac{1}{y^2} =\frac{x^2+y^2}{x^2y^2}$

<u>Let's use what we know and our </u><u><em>useful equation</em></u>:

$(x+y)*(x^2-xy+y^2)*(\frac{x^2+y^2}{x^2y^2} )\\=4*(x^2+y^2-xy)*(\frac{20}{(xy)^2} )\\=4*(20-(-2))*\frac{20}{(-2)^2} \\=4*22*5\\=440$

The value is 440

<u>Answer: 440</u>

Hope that helps!

#LearnwithBrainly

6 0

2 years ago