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Iteru [2.4K]
3 years ago
7

Why dont you square anything with circumference (circles)

Mathematics
1 answer:
rosijanka [135]3 years ago
6 0
Circumference is a distance it is the perimiter of the circle
when you square, you find the legnth and width or the area
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Hello there, <br><br> can someone please help me with this question
monitta

Answer:

10 sides.

Step-by-step explanation:

Two same interior angles + 72 degrees = 360.

Thus, the interior angle of the polygon is \frac{360-72}{2} =144.

The formula to find the interior angle of a polygon is \frac{180(n-2)}{n}

\frac{180(n-2)}{n} =144

Multiply both sides by n:

\frac{180\left(n-2\right)}{n}n=144n

180\left(n-2\right)=144n

Expand:

180n-360=144n

Add 360 to both sides:

180n-360+360=144n+360

180n=144n+360

Subtract 144n from both sides:

180n-144n=144n+360-144n

36n=360

Divide both sides by 36:

\frac{36n}{36}=\frac{360}{36}

n=10

7 0
2 years ago
Given Sin x= 0.5, what is cos x?
Svetllana [295]
Sin x = 0.5
sin x = 1/2
sin x = opp/hyp
therefore the ratio of opp/hyp = 1/2, (opp = 1, hyp = 2)

Find the opp side

1² + x² = 2²
1 + x² = 4
x² = 4 -1
x² = 3
x =√3

The opp side is √3

cos x = opp/hyp = √3/2 = 0.87 (round to the nearest hundredths)
3 0
3 years ago
Given that m∠TSU = m∠SUQ, why is RT || UQ?
o-na [289]
The answer to your question is option A
6 0
2 years ago
Find the missing factor of D that makes the equality true <br><br> -15y^4=(D)(3y^2)<br> D=
sergiy2304 [10]

Answer:

D=-5y^2

Step-by-step explanation:

-15y^4=D*(3y^2)

D=(-15y^4)/(3y^2)=-5y^2

8 0
2 years ago
Find the deriative dy/dx for y=x^2-2x/x^3+3
vaieri [72.5K]

Answer:

\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})

Step-by-step explanation:

So we want to find the derivative of the rational equation:

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

First, recall the quotient rule:

(\frac{f}{g})'=\frac{f'g-fg'}{g^2}

Let f be x^2-2x and let g be x^3+3.

Calculate the derivatives of each:

f=x^2-2x\\f'=2x-2

g=x^3+3\\g=3x^2

So:

\frac{dy}{dx}=(\frac{x^2-2x}{x^3+3})'

Use the above format:

\frac{dy}{dx}=\frac{f'g-fg'}{g^2}\\\frac{dy}{dx}=(\frac{(2x-2)(x^3+3)-(x^2-2x)(3x^2)}{(x^3+3)^2})

And that's our answer :)

(If you want to, you can also expand. However, no terms will be canceled.)

8 0
3 years ago
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