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vlabodo [156]
3 years ago
14

If P is the incenter of JKL, find the measure of JLP

Mathematics
1 answer:
Snezhnost [94]3 years ago
3 0

Answer:

<u>36°</u>

Step-by-step explanation:

> 1/2×180-(32+22)

> 1/2×180-(54)

> 90-54

> 36°

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Compare 6 ⋅ 108 to 3 ⋅ 106.
djverab [1.8K]

Answer:

6*10^8  is 200 times larger than  3*10^6

Step-by-step explanation:

6*10^8\ and \ 3*10^6

To compare both values we divide the exponents

\frac{6*10^8}{3*10^6}

6 divide by 3 is 2

for simplify exponents we use exponential property

a^m / a^n = a^(m-n)

\frac{10^8}{10^6}=10^2

\frac{6*10^8}{3*10^6}=2*10^2= 200

6*10^8  is 200 times larger than  3*10^6

4 0
3 years ago
Read 2 more answers
3:
Klio2033 [76]
M=e/c2

U just divide e=mc2 by c2 on both sides to isolate m
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3 years ago
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FOR A TEST!! HELP! Complete the similarity ratio for the similar triangles.
tatyana61 [14]

Answer:

add them up

Step-by-step explanation:

add 8 10 10 and 80 25 20

7 0
3 years ago
Look at the image please helppp
balu736 [363]

Answer:

Option A. one rectangle and two triangles

Option E. one triangle and one trapezoid

Step-by-step explanation:

step 1

we know that

The area of the polygon can be decomposed into one rectangle and two triangles

see the attached figure N 1

therefore

Te area of the composite figure is equal to the area of one rectangle plus the area of two triangles

so

A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2

step 2

we know that

The area of the polygon can be decomposed into one triangle and one trapezoid

see the attached figure N 2

therefore

Te area of the composite figure is equal to the area of one triangle plus the area of one trapezoid

so

A=\frac{1}{2}(8)(4)+\frac{1}{2}((4+8)(8)=16+48=64\ yd^2

7 0
3 years ago
PLD HELP ME IM STUCK
miv72 [106K]

Answer:

I think it's A: Similar - AA

Step-by-step explanation:

We know that ∠K≅∠N & ∠L≅∠P, so that gives us 2 sets of congruent angles

7 0
3 years ago
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