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stealth61 [152]

2 years ago

13

The perimeter of the base of a square prism is 16cm. The height of the prism is 5 cm. Find the volume of the prism.

1 answer:

zhenek [66]2 years ago

7 0

:$/$$$$677262)2) 677

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stion 13: A typical barrel of bot-oil contains 70 liters of oil. If a hole he barrel is causing a constant 6 liters an hour leak

Montano1993 [528]

Answer:

y = -6x + 70

Step-by-step explanation:

Since the amount of oil in the tank decreases constantly, we need to create a linear function. Using the base form, y = mx + b, we can replace -6 for m because volume decreases by 6 (or increases by -6) each hour, and +70 for b because we start off with 70 liters of oil. So we get y = -6x + 70, with x representing the number of hours, and y representing the volume of oil remaining.

4 0

2 years ago

The Movie House gives away a $5 coupon for every 4 movies purchased.How many movies would you have to purchase in all before rec

Daniel [21]

Answer:

5 movies because for 4 movies equals a coupon

3 0

2 years ago

Please help me I’ll give you a thanks :-)

zloy xaker [14]

Answer:

... sorry

Step-by-step explanation:

...

7 0

2 years ago

According to data from a medical association, the rate of change in the number of hospital outpatient visits, in millions, in

GaryK [48]

Answer:

a) $f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient visits, in millions, is given by:

$f'(t)=0.001155t(t-1980)^{0.5}$

a) To find the function f(t) you integrate f(t):

$\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt$

To solve the integral you use:

$\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}$

Next, you replace in the integral:

$\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C$

Then, the function f(t) is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'$

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient visits.

Hence C' = 264,034,000

The function is:

$f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000$

b) For t = 2015 you have:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

3 0

3 years ago

if possible,find the missing side length labeled with a question mark. if not possible, explain why. on the left at here are 6 i

Lady bird [3.3K]

Answer:

10 inches

Step-by-step explanation:

They're 36 degrees, so corresponding angles are the same, so the middle line is parallel to the bottom line of the triangle.

But the middle line bisects one of the waist lines of the triangle, so this is the middle line, which is equal to half the length of the bottom line

3 0

2 years ago

Read 2 more answers