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Charra [1.4K]
3 years ago
8

Any suggestions on how to solve this? thanks ​

Mathematics
1 answer:
amm18123 years ago
8 0

Answer: x = 4

Step-by-step explanation:

You basically want to solve for the angle A. Do arcsin of 1/3, which is approximately 19.47 degrees. In this triangle, we are missing the adjacent and opposite, but we do have the angle luckily. Lets use SOH-CAH-TOA.

sin(theta) = \frac{opposite}{hypotenuse}

We can see here that doing sine will give us the length of the opposite

sin19.47 = \frac{x}{12}. Do sine of 19.47, then multiply by 12. I got 4

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What is the slope of a line that is perpendicular to the graphed line?
Lemur [1.5K]

Answer:

2

Step-by-step explanation:

To find the slope of the graphed line, you take two points. I have chosen (0, 4) and (8, 0)

The formula of a slope is (y value of second point -y value of first point) divided by (x value of second point -x value of first point):

\frac{0-4}{8-0} =-\frac{1}{2}

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

The slope for a perpendicular line is 2.

6 0
3 years ago
Fern rounded to the nearest ten to estimate 548-132. She subtracted 540-130 and got 410. Is Fern's estimate correct? Explain.
tensa zangetsu [6.8K]
Fern's estimate is not correct because he rounded 548 wrong, it would be rounded to 550 instead since the number 8 is closer to 10. So the correct answer would be 420.
6 0
3 years ago
For the figures below, assume they are made of semicircles, quarter circles and squares. For each shape, find the area and perim
ICE Princess25 [194]

Answer:

Part a) The area of the figure is \frac{9}{2}(4+\pi )\ cm^{2}

Part b) The perimeter of the figure is 3(2+2\sqrt{2}+ \pi)\ cm

Step-by-step explanation:

Step 1

Find the area of the figure

In this problem we have that

The figure ABC is the half of a square and the other figure is a semicircle

<u>Find the area of the figure ABC</u>

we have

AB=6\ cm, BC=6\ cm

The area of the half square ABC is equal to find the area of triangle ABC

so

A1=\frac{1}{2}*6*6=18\ cm^{2}

<u>Find the area of the semicircle</u>

The area of the semicircle is equal to

A2=\pi r^{2}/2

we have that

r=6/2=3\ cm

substitute

A2=\pi (3)^{2}/2

A2=(9/2) \pi\ cm^{2}

The area of the figure is equal to

18\ cm^{2}+(9/2) \pi\ cm^{2}= \frac{9}{2}(4+\pi )\ cm^{2}

Step 2

Find the perimeter of the figure

The perimeter of the figure is equal to

P=AB+AC+length\ CB

we have

AB=6\ cm

Applying Pythagoras theorem

AC=\sqrt{6^{2}+6^{2}}\\AC=6\sqrt{2}\ cm

Remember that

the circumference of a semicircle is equal to

C=\frac{1}{2}2\pi r=\pi r

r=6/2=3\ cm

C=\pi(3)

C=3 \pi\ cm

The perimeter of the figure is equal to

P=6\ cm+6\sqrt{2}\ cm+3 \pi\ cm

Simplify

P=3(2+2\sqrt{2}+ \pi)\ cm

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3 years ago
Use exponents to write 81 three different ways.
nlexa [21]
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A dive ring is 10 feet below the surface of the water in a pool.
USPshnik [31]
100feet below is your answer I think
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