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3 years ago

Find the third side in simplest radical form: 6 √90

Mathematics

1 answer:

Rus_ich [418]3 years ago

6 0

Your answer will be 96 hope this helps :)

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McKenzie bought 1.2 poi d's of coffee for $11.82. what was the cost per pound? help!!

UkoKoshka [18]

When they say 'cost PER pound', the 'per' word means you've to divide cost by pound:

$\frac{cost}{pound} = \frac{11.82}{1.2}$

Doing that division on the calculator will give us $9.85, which is the cost per pound of of coffee!

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4 years ago

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Bonnie baker has a total of 1/2 pound of chocolate. she needs 1/8 pound of chocolate for each batch of brownies. how many batche

ludmilkaskok [199]

1\2-1\8= get the lcm of 2 8=8
(8\2+1=5)(8\8+1=1)
5\8-1\8=4\8=1\2

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4 years ago

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Examine the summary section of the monthly credit card statement below. Use the first five entries to determine whether the new

rjkz [21]

Yea I’m gonna go watch my internet and then go to go home to go to

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2 years ago

Can somebody help me out? I don’t get slope

Shalnov [3]

Answer: the slope of the line is 2/1

Step-by-step explanation:

Find a point on the graph and the go up twice until you see another point on the graph. So go to the right once. So you have found the slope of the line which is 2/1.

Hope this helps ☝️☝☝

5 0

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

7 0

3 years ago