A rectangle has a width of 7 meters and perimeter of 54 meters. What is the length?

Stanley, the skunk, is traveling from point A to point B by taking a shortcut through the pond.A path that avoids the pond is 25

san4es73 [151]

Stanley the skunk will save about 15.5 feet.

This problem can be solve with the Pythagorean Theorem. The possible paths that Stanley can take make a right triangle. The legs of the triangle are 25 and 28, and the hypotenuse would be 37.5.

If Stanley went the long way, he would travel 25 + 28 = 53 feet. Taking the short way (hypotenuse) is 53 - 37.5 = 15.5 feet shorter.

7 0

3 years ago

If David plays 3 tennis matches every week for 9 weeks, how many matches will he play altogether

kari74 [83]

If he plays 3 matches a week for 9 weeks, you have to multiply 3&9.
3*9=27
Daniel will have played 27 tennis matches by the end of the 9 weeks.
Hope this helps:)

5 0

3 years ago

Read 2 more answers

Which if the following items showed the greatest percentage in bulk

Katyanochek1 [597]

No image shown but just remember that percentage is out of 100

0.03 is 3%

7 0

2 years ago

Solve equation using square roots (x-1)^2=4

Aneli [31]

X is equal 3. to do this, you do reverse of order of operations. So this means that you would to the 'times 2' first by dividing 4 by 2 to make it 2. then, in the parenthesis, since there is a 'minus 1', therefore you are going to add 1+2 which equals three. <span />

4 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

3 years ago