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2 years ago

I don't know how to solve this plz help!y = 2 / 3 x - 4

Mathematics

1 answer:

Paha777 [63]2 years ago

8 0

Step-by-step explanation:

-2.666666666 ♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡♡

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You and Mr. Fowles have decided to race each other at the Boston Marathon (26 miles long). Mr. Fowles runs at a steady 4.2 miles

Maslowich

Answer:

The answer is below

Step-by-step explanation:

Speed is the time rate of change of velocity, it is the ratio of distance to time.

Mr. Fowles runs at a steady 4.2 miles per hour. The time taken to cover 26 miles is:

speed = distance / time

time = distance / speed = 26 / 4.2 = 6.19 hour

I run at 3 miles per hour. I was given 3 mile lead, so the remaining distance is 23 miles (26 - 3). The time needed to complete the 23 miles is:

time = distance / speed = 23 / 3 = 7.67 hour

a) Let d be the distance Mr Fowles catches me. Hence:

time = (d + 3)/4.2; also time = d/3

(d + 3)/4.2 = d/3

3d + 9 = 4.2d

1.2d = 9

d = 7.5 miles

He would catch me at 10.5 (7.5 + 3) miles from the starting point.

b) Mr. fowles finishes after 6.19 hour. At that time, the distance is have covered is:

distance = 6.19 * 3 mi/hr = 18.57 mile

Distance behind = 23 mile - 18.57 = 4.43 mile

c) time longer = 7.67 hour - 6.19 hour = 1.48 hour

4 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

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

pounds of salt.

7 0

1 year ago

Someone help pls I’ll mark brainliest if u give a small explanation! :)

marishachu [46]

Answer:

108 inches^2

Step-by-step explanation:

Multiple each side by 1.5

8x1.5=12

6x1.5=9

then to find the area, multiply the sides together.

4 0

2 years ago

Read 2 more answers

1.3n-0.03=-9 (solve for n) grade lvl. 7

bazaltina [42]

Hello from MrBillDoesMath!

Answer:

-6.9

Discussion:

1.3n - 0.03 = -9 => multiply both sides by 100

130n - 3 = -900 =>

130n = -900 +3 = -897 => add 3 to both sides

n = -897/130 =>

n = - (69 * 13)/ (10*13) =>

n = - 69/10 = - 6.9

Thank you,

MrB

4 0

3 years ago

Evaluate. 2−2⋅(12⋅3)−53

klemol [59]

2−2(12)(3)−53
=2−(2)(36)−53
=2−72−53
=−70−53

Answer : −123

3 0

3 years ago

Read 2 more answers

I don't know how to solve this plz help!y = 2 / 3 x - 4 ​

I don't know how to solve this plz help!y = 2 / 3 x - 4