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Lyrx [107]
2 years ago
5

Last week, Akira spent 5 hours doing homework. Brendan did homework for 3/4 as many

Mathematics
1 answer:
yawa3891 [41]2 years ago
8 0

Answer:

3 hours 45 minutes.

Step-by-step explanation: 5 hours is 300 minutes. 3/4 of 300 is 225 minutes. 225 minutes is 3 hours 45 minutes. BAM

Also brainly would be great

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HELP ME <br> solve for the right triangle
VikaD [51]

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent ratios in the right triangle

cos41° = \frac{adjacent}{hypotenuse} = \frac{VW}{VX} = \frac{7}{VX}

Multiply both sides by VX

VX × cos41° = 7 ( divide both sides by cos41° )

VX = \frac{7}{cos41} ≈ 9.3

-----------------------------------------------------------------------

tan41° = \frac{opposite}{adjacent} = \frac{WX}{VW} = \frac{WX}{7}

Multiply both sides by 7

7 × tan41° = WX, thus

WX ≈ 6.1

-------------------------------------------------------------------------

The sum of the 3 angles in a triangle = 180°

Subtract the sum of the given angles from 180° for ∠ X

∠ X = 180° - (90 + 41)° = 180° - 131° = 49°

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3 years ago
Please Help Thanks so much for the help
boyakko [2]

Answer:

The answer is 10

Step-by-step explanation:

7 0
3 years ago
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Me.Woodruff owns an automobile parts store and typically marks up merchandise 32% over warehouse cost. how much would he charge
statuscvo [17]
The answer is 65 so u just leave the house and u go and get drink and hang out with your friends. 
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Expand the expression 10 (n-8)
Tcecarenko [31]
10n-80 is the answer
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3 years ago
a 18 ft tall statue standing next to a globe casts a 12 ft shadow. Of the globe casts a shadow that is 2 ft ling, then how tall
Igoryamba
<h3>Answer:</h3>

3 ft

<h3>Step-by-step explanation:</h3>

The statue's height is 1.5 times the length of its shadow, so we expect the same relationship for the globe.

... 1.5 × 2 ft = 3 ft

_____

<em>Comment on the problem</em>

As a practical matter, with the sun high enough in the sky to cast a shadow shorter than the object's height, it will be quite difficult to measure the length of the shadow of the point at the top of the globe. The shadow of other parts of the globe will interfere.

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3 years ago
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