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3 years ago

60 miles/hour is how many feet/second?

Mathematics

1 answer:

saveliy_v [14]3 years ago

8 0

1 mile = 5280 feet

1 hour = 3600 seconds

60 mil./h = 60 * 5280/ 3600 ft/s = 88 ft/s

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Solve for n -8 = 6+n

bezimeni [28]

Answer:

no solution

Step-by-step explanation:

n -8 = 6+n

Subtract n from each side

n-n -8 = 6+n-n

-8 = 6

This is never true so there is no solution

8 0

3 years ago

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Questions are in file/pictures! <br><br> (as you can see all my questions are from math lol)

77julia77 [94]

Answer:

the first one would be 12 square feet

the second one would be 2961$

and the third one would be 125%

Step-by-step explanation:

i took the test

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3 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

2 years ago

The median of the values:29,24,30,23,28,18,

joja [24]

Answer:

Step-by-step explanation:

18,23,24,28,29,30

(6 in total + 1)/2=3.5.

- the number which is 3.5 is between 24 and 28.

- middle number between these two is 26.

6 0

2 years ago

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Please walk me through how to solve these problems. If you can only solve one that’s fine, I just don’t know where to start. (Th

Wittaler [7]

Answer:

5,-0.5 is the answer to the fist one

Step-by-step explanation:

they want you to give the mid<em>dle </em>point of the line segment when you have graphed the two end points,... find the coordinates of the point the would be in the middle,... Best of luck,... Chow

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3 years ago