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3 years ago

Giving brainliest how to solve area

Mathematics

2 answers:

Harrizon [31]3 years ago

7 0

Answer:

Step-by-step explanation:

It depends on what area you are talking about?

like if you are talking about for a Circle; you use the formula below to Identify the area of your information of a circle;

pi multiplied by the radius of the circle squared.

The formula for a rectangle of an area is below;

A = L * W (Area = Length times Width)

The formula for a square of an area is also below;

First you must find out or look over the information.

Then you must multiply the length of the diagonal's current measurements by themselves.

Now you -divide the number you have by 2.

Okay, now that you divided your number by 2, -again multiply the perimeter by 1/4 to find the length of a side.

stiks02 [169]3 years ago

5 0

To find area you have to multiply the length and the width of the shape.

You might be interested in

What is 3.754×10^5 than round it to the tenth

saul85 [17]

Answer:

375,400

Step-by-step explanation:

3.754*10^5
10^5=100,000
3.754*100,000
375,400

YaYYYY u got it! =)

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20-%20x%20%5E%7B3%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ - x ^{3} } " alt=

noname [10]

I'm guessing you're given the function $y(x)=2-x^3$ , and you're asked to find the inverse function $y^{-1}(x)$ . To do this, swap $x$ and $y$ , then solve for $y$ :

$x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}$

so that the inverse function is

$y^{-1}(x)=\sqrt[3]{2-x}$

Just to verify:

$y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x$

$y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x$

But in case you're actually only interested in computing the square root, first we note that $\sqrt x$ (the real-valued square root) is only defined as long as $x\ge0$ . So $\sqrt{-x^3}$ is defined as long as $-x^3\ge0$ , or $x^3\le0$ , or equivalently $x\le0$ . Under this condition, we could write

$\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}$

We can simplify this further, but we have to be careful. Suppose $x=-1$ . Then $x^2=(-1)^2=1$ . But we get the same result if $x=1$ , since $x^2=1^2=1$ . There are two possible values of $x$ that given the same value of $x^2$ , so to capture both of them, we take $\sqrt{x^2}=|x|$ , the absolute value of $x$ . Then