Ask question

Ask question

All categories

3 years ago

10

What is the unit rate in miles per hour for each runner?

1 answer:

lana66690 [7]3 years ago

4 0

Terrance: 12 miles/1 hour
Jesse: 8 miles/1 hour

You might be interested in

In 4 days, the price of a share of stock rose 3/4 of a point. On average, what was the change in stock price everyday

RoseWind [281]

Answer:

3

Step-by-step explanation:

If you multiply 3/4 and 4 days together you will get 3 so the answer is 3

7 0

3 years ago

Thor invests $750 in an account, with interest compounded continuously. If his investment doubles in value after 9 years, how mu

elixir [45]

9,000 that will be you answer or if is double will be 18,000

5 0

3 years ago

Read 2 more answers

Plz help !!!!!!!!!!!!!!!!!!

podryga [215]

Answer:

i am sorry i do not understand ill try to look harder ill keep u updated

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The area of a parking lot is 600 square meters. A car requires 6 square metes and a bus requires 30

Aleks04 [339]

A=600 with a maximum of 60 vehicles
Each car takes up 6 m
Each bus takes up 30 m

Max amount of cars in the lot would be 600 divided by 6 or 100 cars.

Max amount of buses in the lot would be 600 divides by 30 or 20 buses.

To collect the maximum amount of money using only 60 vehicles, park 20 buses and 40 cars.

4 0

3 years ago

You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then

gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is $d\theta_1 =(14.36p)^o$

Step-by-step explanation:

From the question we are told that

The angle of elevation is $\theta_1 = 15 ^o = \frac{\pi}{12}$

The height of the tree is h

The distance from the base is D

h is mathematically represented as

$h = D tan \theta$ Note : this evaluated using SOHCAHTOA i,e

$tan\theta = \frac{h}{D}$

Generally for small angles the series approximation of $tan \theta \ is$

$tan \theta = \theta + \frac{\theta ^3 }{3}$

So given that $\theta = 15 \ which \ is \ small$

$h = D (\theta + \frac{\theta^3}{3} )$

$dh = D (1 + \theta^2) d\theta$

=> $\frac{dh}{h} = \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta$

Now from the question the relative error of height should be at most

$\pm p$ %

=> $\frac{dh}{h} = \pm p$

=> $\frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta = \pm p$

=> $d\theta = \pm \frac{\theta + \frac{\theta^3}{3} }{1+ \theta ^2} * \ p$

So for $\theta_1$

$d\theta_1 = \pm \frac{\theta_1 + \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} * \ p$

substituting values

$d [\frac{\pi}{12} ] = \pm \frac{[\frac{\pi}{12} ] + \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} * \ p$

=> $d\theta_1 = 0.25 p$

Converting to degree

$d\theta_1 = (0.25* 57.29) p$

$d\theta_1 =(14.36p)^o$

4 0

3 years ago