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4 years ago

This is the last I need help on.

Mathematics

1 answer:

GrogVix [38]4 years ago

7 0

The difference between the x-coordinates of the points given is 3, and none between the y-coordinates, so the side of the square is 3. Square 3 and get the area, 9 units squared.

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Find the value of n in 2m-3n = 6 and 2m + 6n = 12.

Nezavi [6.7K]

Answer:

n = 2/3

Step-by-step explanation:

2m - 3n = 6

– 2m + 6n = 12

-9n = -6

n = -6/-9

n = 2/3

Hence, the value of n in the given equations is 2/3.

4 0

2 years ago

Jessica rides the bus 8% miles each day.how many miles does she ride in 10 days?

liq [111]

Answer:

80 miles

Step-by-step explanation:

3 0

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

JUST 1 PLEASEEEEEEEEEE AND HOW DO YOU KNOW!!!!! PLEASEEEEEEEEEEEEE HELLLPPPPPPPPPPPP MEEEEEEEEEEEEEE

Lemur [1.5K]

Answer:

I can't see three but I'll answer the first two.

Step-by-step explanation:

Number one: The first, third, and fifth options are correct,

Number two: A is correct because if you divide 3/4 you get 0.75. Check by multiplying 0.75 for one by 4 and you get three dollars.

So for three, one pound of pears costs two dollars. That is consistent with the graphs. Pick the answer that states that the best.

5 0

3 years ago

Select the correct answer.

NISA [10]

Answer: 2 aka B

Step-by-step explanation: the answers B :)

3 0

3 years ago