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Anna35 [415]
3 years ago
15

The length, l, that a spring is stretched varies directly as the amount of force, F, applied to the end of the spring. When a fo

rce of 10 pounds is applied to a spring, it stretches 3 inches. How much will the same spring stretch if the force is increased to 15 pounds?
A.
8 inches
B.
7.5 inches
C.
4.5 inches
D.
2 inches
Mathematics
1 answer:
Elina [12.6K]3 years ago
7 0

Answer:

C. 4.5 inches

Step-by-step explanation:

Given the following data;

Force, F = 10 pounds

Extension, e = 3 inches

First of all, we would determine the spring constant (k) using the following formula;

Force = spring constant * extension

10 = spring constant * 3

Spring constant = 10/3

Spring constant = 3.33 pounds per inches.

Next, we would find the extension of the spring when a force of 15 pounds is applied;

F = ke

15 = 3.33 * e

Extension, e = 15/3.33

Extension, e = 4.5 inches.

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<h3>Coordinate plane</h3>

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8 0
3 years ago
Please find the domain and range!
SpyIntel [72]

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Range: {-5, 1, 7, 9}

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{(6, -5), (-2, 9), (-3, 1), (10, 7), (8, 9)}

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This includes all the set of the x-values that are in the relation. This includes, 6, -2, -3, 10, and 8.

Thus, the domain can be represented as:

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3 0
3 years ago
Score Frequency
RUDIKE [14]

Scores on an AP Exam at a local high school are recorded in the table. What was the average score?

Answer: D) 4.266 is correct

Explanation:

We are given the below frequency table:

Score Frequency

1         |         0

2        |         3

3        |         4

4        |        16

5        |       22

Let's denote the score by x and frequency by f.

The average formula is:

Average=\frac{\sum fx}{\sum f}

                    =\frac{(1\times0) + (2\times3)+(3\times4) +(4\times16) + (5\times22)}{0+3+4+16+22}

                    =\frac{0+6+12+64+110}{45}

                    =\frac{192}{45}=4.267

Therefore, the average score was 4.267

Hence the option D) 4.266

7 0
4 years ago
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