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andrew-mc [135]
3 years ago
6

Tessie has a volleyball game at 6:45 p.m. she needs to be there 15 minutes early to warm up for the game and it takes her 40 min

utes to go to The gym what time should she leave her house
Mathematics
1 answer:
const2013 [10]3 years ago
6 0
First we need to add the time she takes warming and the time she takes going to the the gym:
15minutes+40minutes=45minutes

Next, we area going to subtract those 45 minutes from the time the game begins: 
6:45pm-45minutes=6:00pm

We can conclude that she should leaver her house at 6:00 pm if she wants to get in time for her 6:45 pm game.
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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
Which value of c makes the ratios equal? *
Nikitich [7]

Answer:

Option 4

The answer is 10.8

Step-by-step explanation:

5/6 = 9/c

5c = 9 × 6

5c = 54

5c/5 = 54/5

c = 10.8

Thus, The value of c is 10.8

<h3><u>For </u><u>Verification</u>;</h3>

5/6 = 9/c

5/6 = 9/10.8

0.833 = 0.833

Hence, L.H.S = R.H.S

<u>-TheUnknownScientist</u><u> 72</u>

3 0
2 years ago
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