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3 years ago

8

If the new building is 22.5 inches from the gymnasium in the model, what would the actual distance from the gymnasium to the new

building be? INCLUDE UNITS!
1 inch = 3 feet

2 answers:

Rufina [12.5K]3 years ago

8 0

67.5 =)
hope this helps

Klio2033 [76]3 years ago

4 0

Answer:

yes

Step-by-step explanation:

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Furkat [3]

Answer:

Around 0.22

<u>FIRST THOUGH, I FEEL LIKE 90 MIGHT HAVE MEANT TO BEEN 0.9, SO IF SO SUBSTITUTE THE 90 BELOW FOR 0.9!</u>

<u></u>

Step-by-step explanation:

Okay, so first.

Company A will be 50 a day plus 0.4 a mile

Company B will be 30 a day plus 90 per mile

We can write an equation like this:

50 + 0.4m = 30 + 90m

m = miles they are the same

Then we solve for m.

50 + 0.4m = 30 + 90m

- 0.4m - 0.4m

50 = 30 + 89.6m

- 30 - 30

20 = 89.6m

Divide both sides by 89.6

Around 0.22!

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3 years ago

*****50 POINTSSSS*****

defon

Answer:

<h3>Given</h3>

m∠REG = 78°
mAR = 46°
ER ≅ GA

<h3>Solution</h3>

m∠GAR = 180° - m∠REG = 180° - 78° = 102° (supplementary angles sum to 180°)
m∠TAR = 1/2mAR = 1/2(46°) = 23° (tangent chord angle is half the size of intercepted arc)
m∠GAN = 180° - (m∠TAR + m∠GAR) = 180° - (23° + 102°) = 55° (straight angle is 180°)
mAG = 2m∠GAN = 2(55°) = 110°
mRE = mAG = 110° (as ER ≅ GA)
mGE = 360° - (mAG + mAR + mRE) = 360° - (110° + 46° + 110°) = 94° (full circle is 360°)

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Answer:

Step-by-step explanation:

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