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4 years ago

Help me on 13 please I don't get it

Mathematics

2 answers:

zubka84 [21]4 years ago

8 0

5, 5, regular polygon. I think.

Georgia [21]4 years ago

3 0

13a: the figure has 5 sides and 5 vertices since it is a hexagon.

13b: is not a regular polygon.
It is a polygon, but it isn't a regular polygon like a square is.

I hope this helps:)

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How do I turn this 10(0j + 40 + 4j) into a verbal expression? <br> I NEED HELP!!!!1

egoroff_w [7]

Answer:

See below

Step-by-step explanation:

You could say "10 times the sum of 0j, 40, and 4j" because that is exactly what you are doing in the algebraic expression.

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2 years ago

What is 40/100 Lowest Term

sergejj [24]

$\frac{40}{100}=\frac{2 \times 20}{5 \times 50}=\boxed{\frac{2}{5}}$

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4 years ago

What is the answer to this equation <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7B4%7D%3D5%20" id="TexFormula1" title="

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4 years ago

An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as

Paul [167]

Answer:

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is $\frac{a}{\sqrt{3}}=R$ .

Upon substituting our given values, we will get:

$\frac{5x}{\sqrt{3}}=6r$

Let us solve for r.

$r=\frac{5x}{6\sqrt{3}}$

$\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}$

We know that area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4}s^2$ , where s represents side length of triangle.

$\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2$

The area within circle and outside the triangle would be difference of area of circle and triangle as:

$A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}$

We can make a common denominator as:

$A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}$

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Therefore, our required expression would be $A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$ .

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3 years ago

What are the coordinates of the center of the circle whose equation is (x – 9)2 + (y + 4)2 = 1?

fiasKO [112]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x-9)^2+(y+4)^2=1
\\\\\\\
[x-\stackrel{h}{9}]^2+[y-(\stackrel{k}{-4})]^2=1\qquad \qquad \qquad center~(9,-4)$

6 0

4 years ago

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