PLZ HELPP 10 POINTS PLZZZ

Fudgin [204]

51 ft is the answer

Explanation

The hight of the man divided by the hight of school ( which is unknown )

Equal

The length of the shadow of the man divided by the shadow of the school

Cross multiplication and find the value of unknown x which is 51... and ya

4 0

2 years ago

In the line 3y=4x-1 and 9y=x+3 are parallel to each other the value of q is

jonny [76]

Do you mean the value of y? there is no q in these equations.

5 0

2 years ago

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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

1 year ago

What is the surface area of the right cylinder below. radius 3 height 7

lina2011 [118]

Answer:

160 sq. uits

Step-by-step explanation:

160 sq. uits é a resposta

4 0

2 years ago

Solve by Substitution<br> Show Steps<br> −2x + 3y + 5z = −21<br> −4z = 20<br> 6x − 3y = 0

Anarel [89]

−2x + 3y + 5z = −21

−4z = 20

6x − 3y = 0

do -4z=20 first

divide both sides by -4 to get z by itself

-4z/-4=20/-4

z=-5

Use z=-5 into −2x + 3y + 5z = −21

-2x+3y+5(-5)=-21

-2x+3y-25=-21

move -25 to the other side

sign changes from -25 to +25

-2x+3y-25+25=-21+25

-2x+3y=4

6x-3y=0

find x by eliminating y

Add the equations together

-2x+6x+3y+(-3y)=4+0

-2x+6x+3y-3y=4

4x=4

Divide by 4 for both sides

4x/4=4/4

x=1

Use x=1 into 6x − 3y = 0

6(1)-3y=0

6-3y=0

Move 6 to the other side

6-6-3y=0-6

-3y=-6

Divide both sides by -3

-3y/-3=-6/-3

y=2

Answer:

(1, 2, -5)

5 0

3 years ago

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