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givi [52]
2 years ago
10

PLS HELP ASAP!!! In the following diagram DE | FG and KL | FG what is the measure of angle X

Mathematics
1 answer:
myrzilka [38]2 years ago
6 0

Answer:

<x=28 degrees

Step-by-step explanation:

Since we know line KL and FG are perpendicular, and DE and FG are parallel, then KL and DE must also be perpendicular.

Angles GCA, JCB, IAE, and CAD are all the same angle. Angles IAD, EAC, GCJ, and ACB are all the same angle. Since DE and KL are perpendicular, then angles KAD, KAE, EAL, and DAL all are 90 degrees, or right angles.

Because we know all this, we can take the angle of KAD, 90 degrees, and subtract 62 from it to find the angle of IAD. This is 28 degrees. Since angle IAD is equal to angle GCJ, which is also angle x, angle x is 28 degrees.

Hope this helps! Please mark me Brainliest!

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3 years ago
In triangle NOP p equals 600 inches angle P equals 96° and angle N equals 64° what is the area of triangle NOP to the nearest 10
DochEvi [55]

9514 1404 393

Answer:

  55,637.8 square inches

Step-by-step explanation:

We can find side n using the Law of Sines:

  n/sin(N) = p/sin(P)

  n = p(sin(N)/sin(P)) = 600·sin(64°)/sin(96°)

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The angle O is ...

  O = 180° -N -P = 180° -64° -96° = 20°

Then the area is ...

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  A = (1/2)(542.246913 in)(600 in)·sin(20°) ≈ 55,637.81008 in²

The area of ∆NOP is about 55,637.8 in².

7 0
3 years ago
Please help in math I give lots of points.
disa [49]

\text{Put the value of x to the expression}\\\\\sqrt{x+3};\ x=15\\\\\sqrt{15+3}=\sqrt{18}=\sqrt{9\cdot2}=\sqrt9\cdot\sqrt2=\boxed{3\sqrt2}


\text{Put the value of x to the expression}\\\\\dfrac{6}{\sqrt{x}},\ x=3\\\\\dfrac{6}{\sqrt3}=\dfrac{6\cdot\sqrt3}{\sqrt3\cdot\sqrt3}=\dfrac{6\sqrt3}{3}=\boxed{2\sqrt3}


\text{Put the value of x to the expression}\\\\4\sqrt{x+1}-3\sqrt{2x+6},\ x=3\\\\4\sqrt{3+1}-3\sqrt{2(3)+6}=4\sqrt4-3\sqrt{6+6}=4(2)-3\sqrt{12}\\\\=8-3\sqrt{4\cdot3}=8-3\sqrt4\cdot\sqrt3=8-3(2)\cdot\sqrt3=\boxed{8-6\sqrt3}

8 0
3 years ago
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