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netineya [11]
3 years ago
9

In ΔSTU, u = 40 inches, ∠S=73° and ∠T=71°. Find the length of t, to the nearest inch.

Mathematics
1 answer:
zhuklara [117]3 years ago
5 0

Answer: 64 inches

Step-by-step explanation: Solve using sohcahtoa for missing info

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The drawing is a scale drawing of a house.
Alborosie

Answer: 20.25ft

1. Find what 8 inches in scale equal for the height of the house.

we will do this step by multiplying.

2in=4.5

8=?

to find...

2x4=8

so

4.5x4=18

2. Find what 1 inch on the scale is equal to for the height of the house.

we will do this by dividing 4.5 by 2

4.5/2=2.25

3. Add both the values for step 1 and 2.

we will do this by just adding them both

18+2.25=20.25

<h2>Why is this our answer?</h2>

This is our answer, because first we found the value of 8 inches on the scale, which means that we are finding almost the full value of the 9 inches of the height of the house, then, we found 1 inch because 9-8=1, so if we already found 8, which is a number in the table of 2, we found out 1 so that we can add it with 8 to find the value of 9in on the scale. In this way, after adding, that gives us the value of 9in on the scale!

My gratitude attitude - THANKS!

3 0
2 years ago
Simplify the expression. 3.7z to the power of 2+7.2+9z-1.2-2z+z to the power of 2.
Musya8 [376]
The answer is Would be 2 hehehehe hshshs x=2
4 0
3 years ago
Spencer bought a soccer ball for $9. 95 and z baseballs for $2. 99 each. Which of the following expressions shows the total amou
Elan Coil [88]

Answer:

9.95 + 2.99z

Step-by-step explanation:

9.95 + 2.99z

6 0
3 years ago
Who is the richest president in the world​
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The richest president in the world is Sebastian Pinera, President of Chile with $2.4 billion.

3 0
3 years ago
What curve passes through the point ​(1 ​,2​) and has an arc length on the interval​ [2,6] given by Integral from 2 to 6 StartRo
Georgia [21]

Answer:

Step-by-step explanation:

Given

Length of curve

L=\int_{2}^{6}\sqrt{1+64x^{-6}}dx

Length of curve is given by

L=\int_{a}^{b}\sqrt{1+\left ( \frac{\mathrm{d} y}{\mathrm{d} x}\right )^2}dx over interval a to b

comparing two we get

\frac{\mathrm{d} y}{\mathrm{d} x}=8x^{-3}

dy=8x^{-3}dx

integrating

\int dy=\int 8x^{-3}dx

y=-4x^{-2}+C

Curve Passes through (1,2)

1=-4+C

C=5

curve is

y+\frac{4}{x^2}=5

3 0
3 years ago
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