The formula for acceleration is (∆v)/(∆t), or (final velocity - initial velocity)/(final time - initial time). If you input your values the problem would work out as followed:

A = <span>(∆v)/(∆t)
A = (25-9)/(8-5)
A = 16/3
A = 21.33 (rounded to the nearest hundredth)</span>

8 0

3 years ago

What are the solutions of the equation x^4-5x^2-36=0? Use factoring to solve

Murljashka [212]

X = -3, 3, 2i, -2i
Hope this helps! :)

3 0

3 years ago

What is the value of x ?

Naddika [18.5K]

Answer:

x = 108°

Step-by-step explanation:

x + 2/3 x = 180°

1 2/3x = 180°

5/3 x = 180°

180 * x = 540

540 / 5 = 108

x = 108°

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

6 0

3 years ago

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along

nordsb [41]

Answer:

20/3 ft/sec

Step-by-step explanation:

As given the diagram (attached),

x is the distance from the man to the pole, and y is the distance from the tip of the man's shadow to the pole.

I assume that the man and the pole are upright, meaning the 2 triangles are identical.

By similar triangles, we will get

15/y = 6/(y-x)

Triangles are similar if they have the same shape, but can be different sizes. In our case, triangles are in similar shape but in different sizes

15 (y-x) = 6y

15y-15x=6y

9y=15x

y=5/3 x

As we need to find rate of change (How fast), differentiate both sides with respect to time t. we get

dy/dt = 5/3 dx/dt

As we have given that the man is walking from the with a speed to 4ft/s

so dx/dt = 4, So, we need to find dy/dt (How fast the tip of shadow moving)

dy/dt = 5/3 *4

dy/dt = 20/3 or 6.666 ft/sec

In this case, the man's distance from the pole doesn't matter, because only his velocity influences how quickly his shadow goes.

5 0

3 years ago