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bearhunter [10]

3 years ago

8

Please tell me if I'm right or not

1 answer:

Pepsi [2]3 years ago

8 0

It’s actually 25.2% of boys prefer to play inside

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PLEASE HELP ;)<br> Who did it right?<br><br> Find the slope of the line perpendicular to 2x+3y=8

levacccp [35]

Answer:

Marissa.

Step-by-step explanation:

To find perpendicular slope you basically take the slope of the given line and reverse whatever sign it had previously. (If it was negative, it would become positive. If it was positive it would become negative.)

By doing -A/B it make the previous slope negative.

Brandon only flipped the fraction which would just result in the slope being slightly steeper, or in this case, less steep than the previous line, rather than making it perpendicular.

Hope this helps! Stay safe and if you have any questions feel free to ask!

6 0

2 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

An olympian swam the 200-meter freestyle at a speed of 1.8 meters per second. An olympic runner ran the 200-meter dash in 21.3 s

Maurinko [17]

Answer:

7.6m/s

Step-by-step explanation:

Find the time the Olympian swam

Speed=distance/time ⇒⇒Time=distance/speed

Distance=200m , speed= 1.8m/s t= 200/1.8 = 111.11 seconds

Find the speed of the Olympic runner

Distance=200m time = 21.3 sec

s=200/21.3 = 9.4 m/s

Difference in speed= 9.4-1.8= 7.6m/s

5 0

3 years ago

Read 2 more answers

Katie has a 45% chance of making an A on her Algebra II test and a 55% chance of making an A on her economics test. What is the

gogolik [260]

The answer would be D) 50%

7 0

2 years ago

Read 2 more answers

What is two other ways to write 8.517

vova2212 [387]

One way is in word form the other way is in fraction form example

1. eight and five-hundred seventeen thousandths
2. 8517/1000

6 0

2 years ago