The graph below shows the force required to stretch a spring various distances. Review the graph below. A graph of Force versus Elongation. Elongation (m) is on the x axis and Force (N) is on the y axis. A straight line passes through the points 0 comma 0 and 10 comma 15.

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters? (2 points)

Group of answer choices

4.5 J

6.8 J

7.5 J

14 J

step-by-step explanation:

in which figure is point g a centroid

-8/2y-8=5/y+4-7y+8/y^2-16

we move all terms to the left:

-8/2y-8-(5/y+4-7y+8/y^2-16)=0

we can not solve this equation

the value of x is 9 ⇒ answer b

step-by-step explanation:

* we can solve this problem using cosine and sine rule

- in δ abc

∵ ab = 36 , bc = 28 , ac = 16

- lets find the measure of angle b using the cosine rule

∵ cos b = [ab² + bc² - ac²]/2(ab)(bc)

∴ cos b = [36² + 28² - 16²]/2(36)(28)

∴ cos b = 19/21

∴ m∠b = cos^-1(19/21) ≅ 25°

- lets find the measure of angle a using the sine rule

∵ sina/bc = sinb/ac = sinc/ab

∴ sina/28 = sin25°/16 ⇒ by using cross multiplication

∴ sina = 28 × sin25°/16

∴ sina = 0.7396

∴ m∠a = sin^-1(0.7396) ≅ 48°

- from the figure m∠abd = m∠dbc

∵ m∠abc = 25°

∴ m∠abd = 25°/2 = 12.5°

- in δabd

∵ m∠a = 48° , m∠abd = 12.5°

∵ the sum of the measures of the interior angles of any δ is 180°

∴ m∠adb = 180° - (48° + 12.5°) = 119.5°

∵ ab = 36

∵ ad = x

- use the sine rule to find x

∵ sin∠adb/ab = sin∠abd/x

∴ sin119.5°/36 = sin12.5°/x ⇒ by using cross multiplication

∴ x = 36 × sin12.5°/sin119.5° = 8.95 ≅ 9

* the value of x is 9