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yawa3891 [41]
3 years ago
14

Two cylinders, a and b, each started with different amounts of water. The graph shows how the height of the water changed as the

volume of water increased in each cylinder. Which cylinder has a larger radius? Explain how you know.
Mathematics
2 answers:
Sauron [17]3 years ago
7 0

Answer:

:)

Step-by-step explanation:

Elis [28]3 years ago
5 0

Answer:

There is no graph

Step-by-step explanation:

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1 quarter 7 dimes and 23 pennies in decimal and fraction form
Lerok [7]

Answer:

Thanks for asking!

Step-by-step explanation:

The answer would be 1.18 in decmial form and 1 9/50

3 0
2 years ago
What is the value of x in the equation −8 + x = −2? (1 point)
GREYUIT [131]
The answer is positive 6
4 0
3 years ago
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Over 6 days jim jogged 6.5 miles 5 miles 3 miles 2 miles 2 miles 3.5 miles and 4 miles. What is the mean distance that jum jogge
Marizza181 [45]

Mean means average.

To find the average, add the data then divide it by the amount of individual data you have.

(6.5 + 5 + 3 + 2 + 2 + 3.5 + 5)/2 = mean

27/2 = mean

13.5 = mean

13.5 miles is the mean distance that Jim jogged.

5 0
3 years ago
Find the area if the shaded segment. Round your answer to the nearest square foot.
8090 [49]
2 ft I think
Is the answer
5 0
3 years ago
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In ΔJKL, the measure of ∠L=90°, JL = 24, LK = 7, and KJ = 25. What is the value of the sine of ∠K to the nearest hundredth?
podryga [215]

<u>Given</u>:

Given that JKL is a right triangle.

The measure of ∠L is 90°

The length of JL = 24, LK = 7, and KJ = 25.

We need to determine the value of sine of ∠K.

<u>Value of sin ∠K:</u>

The value of sin ∠K can be determine using the trigonometric ratio.

Thus, we have;

sin \ \theta=\frac{opp}{hyp}

Substituting \theta=K, the side opposite to angle K is JL and the hypotenuse is JK

Thus, we have;

sin \ K=\frac{JL}{JK}

Substituting JL = 24 and JK = 25, we get;

sin \ K=\frac{24}{25}

Simplifying, we get;

sin \ K=0.96

     K=sin^{-1}(0.96)

     K=73.74^{\circ}

Thus, the measure of sine of ∠K is 73.74°

6 0
3 years ago
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