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pickupchik [31]
3 years ago
14

at the playground, the new sandbox was 3 meters wide and 2 meters long. what is the area of the sandbox

Mathematics
2 answers:
77julia77 [94]3 years ago
8 0
For any rectangle (like a sandbox) the area is equal to the length times the width. The length of this rectangle is 2 and the width is 3, so the area is equal to 2 m*3 m=6 square meters
S_A_V [24]3 years ago
3 0

area = length times width

area = 2 meters * 3 meters

area = 6 m^2

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Please help me for the love of God if i fail I have to repeat the class
Elena-2011 [213]

\theta is in quadrant I, so \cos\theta>0.

x is in quadrant II, so \sin x>0.

Recall that for any angle \alpha,

\sin^2\alpha+\cos^2\alpha=1

Then with the conditions determined above, we get

\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35

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\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}

Now recall the compound angle formulas:

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta

\sin2\alpha=2\sin\alpha\cos\alpha

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as well as the definition of tangent:

\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}

Then

1. \sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}

2. \cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}

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\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2

\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2

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