Consider the relationships labeled in the diagram. Q Á P Р R Find m

Licemer1 [7]

The answer is A. Even though it have a positive 3, the horizontal shift is opposite of what it is

4 0

3 years ago

jeremy puts on empty container under a leaking faucet. in the first hour 3/8 quart of water collects. in the second hour, 1/6 qu

QveST [7]

Answer:

13/24

Step-by-step explanation:

=3/8+1/6=9/24+4/24=(9+4)/24=13/24

7 0

4 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

A patio is to be made in the shape of an isosceles triangle as shown.

ira [324]

9514 1404 393

Answer:

(a) 22 is not the correct value of h

(b) 164 ft²

(c) $2009

Step-by-step explanation:

(a) Apparently, the formula ...

A = 1/2bh

was used. We recognize the base of the triangle is 16 feet, so apparently the formula was used with h=22.

22 is the side length, not the perpendicular distance from the base to the opposite vertex.

The computed value is not the area because 22 is not the height.

__

(b) Given three side lengths, an appropriate formula is Heron's formula:

s = (a+b+c)/2 and A = √(s(s-a)(s-b)(s-c))

Here, we have ...

s = (22 +22 +16)/2 = 30 and A = √(30(8)(8)(14) = 16√105 ≈ 163.95

The area is about 164 square feet.

__

(c) At $12.25 per square foot, the cost of materials will be ...

(164 ft²)($12.25/ft²) = $2009

8 0

3 years ago

A ship left shore and sailed 240 kilometers east, turned due north, then sailed another 70 kilometers. How many kilometers is th

Elza [17]

Your Answer will be:

7 0

3 years ago