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2 years ago

Find the measurement of angle A

Mathematics

1 answer:

yaroslaw [1]2 years ago

6 0

Answer:

The value of A is 20

Step-by-step explanation:

To find the answer, note that all triangles have a interior degree measurement of 180. Set the sum of the angles equal to 180.

71x - 1 + 20x + 90 = 180

91x + 89 = 180

91x = 91

x = 1

Now we can use this in the A angle to find the value.

A = 20x

A = 20(1)

A = 20

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how many times as much is the number of students at Pengrove university has the number of students at Allentown College?

svp [43]

The answer would be: D-100. The reasoning of this is because if you take 3 X 10 to the third power and did the exponent (10 to the third power in this case) and multiplied it you would obviously get 1,000 for an answer. Then would would do 1,000 X 3 to get 3,000.

5 0

2 years ago

Plz answer!! only if you are correct. don’t answer just for points :)

Ann [662]

Answer

Step-by-step explanation:

Both 1 and 2 equel up to 90 degrees

so to find x for 1 and 2 you would simplify the equations

90 = 6x - 3

and

90= 3x - 6

simplify them and you wil get the answers you want

another way to do it is combining the two and solve.

90 = 9x - 9

Just simplify :)

8 0

2 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

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