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motikmotik
3 years ago
14

Shane wrapped the kite with paper. How many square inches of paper did Shane use to cover the front and back? 45 POINTS

Mathematics
2 answers:
arsen [322]3 years ago
7 0

Answer:

576

Step-by-step explanation:

First off, we can use the area of a kite formula, (D1*D2)/2 to get the area of one side of the kite, which is simplified to 18 * 32 : which gets us 288. But because we are looking for the area of the front and back side, we multiply 288 by 2- getting the final answer of <u>576</u>.

meriva3 years ago
5 0

Answer:

576

Step-by-step explanation:

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I like turtles so how fast can a turtle cross the street if it was going 2 mils per hour and 2,934,000 hours past?
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Answer:

5,868,000

Step-by-step explanation:

2,934,000 x 2 = 5,868,000

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Find the value of f(3) for the function.<br> f(x) = -2(x + 3)
BabaBlast [244]

Answer:

-2(3+3) = -12

Step-by-step explanation:

f(x) is telling you to substitute all the x values into the equation. The f(x) of the function is 3, telling you to plug in 3 in the equation. This leads to -2(6) = -12.

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Alinara [238K]

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92

Step-by-step explanation:

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3 years ago
What value of x makes the equation -2/3x + 3 1/3x - 1/2 = -1/3x + 5 1/2 true? Please help ASAP!!!!!!!!
Naddika [18.5K]

Answer:

<h2>x = 2</h2>

Step-by-step explanation:

<h2 />

The steps of the solution in the photo above

I hope that is useful for you :)

3 0
3 years ago
Find the exact value of each trigonometric function for the given angle θ.
Kay [80]

Answer:

\sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=\dfrac{2}{\sqrt{3}}.

Step-by-step explanation:

The given angle is 240 degrees.

We need to find the exact value of each trigonometric function for the given angle θ.

Since \theta=240, it means θ lies in 3rd quadrant. In 3d quadrant only tan and cot are positive.

\sin (240^\circ)=\sin (180^\circ+60^\circ)=-\sin (60^\circ)=-\dfrac{\sqrt{3}}{2}

\cos (240^\circ)=\cos (180^\circ+60^\circ)=-\cos (60^\circ)=-\dfrac{1}{2}

\tan (240^\circ)=\tan (180^\circ+60^\circ)=\tan (60^\circ)=\sqrt{3}

\cot (240^\circ)=\cot (180^\circ+60^\circ)=\cot (60^\circ)=\dfrac{1}{\sqrt{3}}

\sec (240^\circ)=\sec (180^\circ+60^\circ)=-\sec (60^\circ)=-2

\csc (240^\circ)=\csc (180^\circ+60^\circ)=-\csc (60^\circ)=-\dfrac{2}{\sqrt{3}}

Therefore, \sin (240^\circ)=-\dfrac{\sqrt{3}}{2},\cos (240^\circ)=-\dfrac{1}{2},\tan (240^\circ)=\sqrt{3},\cot (240^\circ)=\dfrac{1}{\sqrt{3}},\sec (240^\circ)=-2,\csc (240^\circ)=-\dfrac{2}{\sqrt{3}}.

8 0
3 years ago
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