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Zielflug [23.3K]
2 years ago
5

Please help me answer ASAP

Mathematics
1 answer:
Vsevolod [243]2 years ago
6 0

Answer: 4.5

Step-by-step explanation:

\sin 33^{\circ}=\frac{\sqrt{6}}{CE}\\CE \sin 33^{\circ}=\sqrt{6}\\CE=\frac{\sqrt{6}}{\sin 33^{\circ}} \approx \boxed{4.5}

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about 16°

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Make sure your calculator is in degree mode.

Drawing the right triangle, we see that the shorter leg is 34 meters, and the longer leg is 122 meters. So the angle of elevation is:

\tan( \alpha )  =  \frac{34}{122}

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Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if
Bezzdna [24]

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using distance formula, we get

JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}

JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}

JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}

JK=\sqrt{4+9}

JK=\sqrt{13}

Similarly,

KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6

LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}

JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6

Now, perimeter P of ▱JKLM is

P=JK+KL+LM+JM

P=\sqrt{13}+6+\sqrt{13}+6

P=2\sqrt{13}+12

P=2(3.61)+12

P=7.22+12

P=19.22

P\approx 19.2

Therefore, the perimeter P of ▱JKLM is 19.2 units.

3 0
3 years ago
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