All categories

3 years ago

A wheel has a diameter of 3.5ft. Approximately how far does it travel if it makes 20 complete revolutions? use 22/7 for pi.

1 answer:

8 0

If the wheel doesn't slip or skid, then every time it makes
a revolution, it travels the circumference of the wheel.

If its diameter is 3.5 ft, then its circumference is (3.5 pi)

= (3.5 ft) x (22/7) = 77/7 ft = 11 ft .

In 20 complete revolutions, it travels

(20) x (11 ft) = 220 ft .

You might be interested in

If my assignment is worth 14% of my quiz weight which is 27% of my total weight what is the weight of that assignment

Nimfa-mama [501]

Answer:

3.785%

Step-by-step explanation:

Assignment is 14% of the quiz

=> Assignment = 14% of the quiz

Quiz is 27% of the total weightage

=> 14% of 27%

14% of 27%

= 0.14 x 27 = 3.785%

The assignment is 3.785% of the total weightage.

6 0

3 years ago

How to do this question..?

Leokris [45]

uh idk um...

Okay well to answer this you have to work the problem out and enable to do that you have to look at the information provided, also STOP CHEATING AND FIGURE IT OUT

um i mean.....GOODLUCK!!!! I know you can do it :)

7 0

3 years ago

David can run at a consistent rate of 8.2 miles per hour. If he has 3 and a half hours to

Nataliya [291]

Answer:

David can run 28.7 miles in 3.5 hours.

Step-by-step explanation:

8.2mph × 3.5hours = 28.7miles

<em>good luck, i hope this helps :)</em>

7 0

3 years ago

Read 2 more answers

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago

Find x in the given figures. x = inches. _√_

Delvig [45]

Answer:

X = 4√3

Step-by-step explanation:

Using tangent-secant theorem

So,

X² = (4)(4+8)

X² = (4)(12)

X² = (48)

Taking sqrt on both sides

X = 4√3

8 0

3 years ago