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Misha Larkins [42]
3 years ago
8

The approximate radius of the Earth is 6371 km. The approximate radius of the Moon is 1737.5km. What is the difference in their

circumferences
Mathematics
1 answer:
oee [108]3 years ago
6 0

Answer:

earth C≈40030.17

moon C≈10917.03

the difference in their circumference is 29113.14 km

Step-by-step explanation:

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Complete the following statements to prove that ∠IKL and ∠JLD are supplementary angles. It is given that ∠EIJ ≅ ∠GJI. Also, ∠EIJ
Dima020 [189]

Answer:

1st blank: substitution property of equality

2nd blank: linear pair theorem

3rd blank: substitution property of equality

Step-by-step explanation:

<u>1st blank</u>

∠EIJ ≅ ∠GJI (eq. 1)

∠EIJ ≅ ∠IKL (eq. 2)

∠GJI ≅ ∠JLK (eq. 3)

Substituting eq. 3 into eq. 1:

∠EIJ ≅ ∠JLK

and then, substituting eq. 2:

∠IKL ≅ ∠JLK

which means that m∠IKL = m∠JLK

<u>2nd blank</u>

The Linear Pair Theorem states that two angles that form a linear pair are supplementary

<u>3rd blank</u>

m∠JLK + m∠JLD = 180°

Substituting with the previous result:

m∠IKL + m∠JLD = 180°

6 0
3 years ago
s:Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar and ^ to indicate
anzhelika [568]
Vco = 1/3×πr2h
Vcy = Vco --> πr^2×h = 1/3πR^2×h
h's cancel, as well as π's
r^2 = 1/3R^2 --> R^2 = 3r^2 -->
SR (R^2) = SR (3r^2)
R = SR(3)×r
8 0
3 years ago
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If nobody answers this they think im hot
cupoosta [38]

NO YOU ARE NOT

AND THANK YOU

5 0
3 years ago
Pls solve urgent will mark branliest
Natalka [10]

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

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8 0
2 years ago
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What is the cost for 1.4lb item that is $1.58 per pound?
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