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2 years ago

2) A jacket that is regularly priced at $175 is on sale, at a discount of 35%.

Mathematics

1 answer:

Mandarinka [93]2 years ago

5 0

Answer:

$113.25

Step-by-step explanation:

35% of 175 is 61.25. Since it's 35% OFF we have to subtract 61.25 to 175 which is 113.25.

I hope this helps and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

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Grace rides her bike 21 km in 1 1/6 hours. What is her average speed in kilometers per hour?

aleksandrvk [35]

126 because 20 X 6 and 6 x 1

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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

The burges are moving across the country .Mr burges leaves 3 hours before mrs.burges. If he averages 50 mph and she averages 70

kirill [66]

10.5 hours

let t be time then distance travelled by both is

Mr Burges ⇔ d = 50t ( distance = speed × time )

Mrs Burges ⇒ d = 70(t - 3)

At the point of overtaking both will have travelled the same distance

equating their distance and solving for t

70(t - 3) = 50t

70t - 210 = 50t ( subtract 50t from both sides )

20t - 210 = 0 ( add 210 to both sides )

20t = 210 ( divide both sides by 20 )

t = 10.5 hours

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3 years ago

Whoever answers this correctly gets brainliest

ELEN [110]

Answer: 11. Is A

Step-by-step explanation:

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3 years ago

What is the equation of this line

Oxana [17]

Answer:

y = x + 2

If you use y = x, it will become in the direction of that line. I am assuming 1 block = 1 unit of measurement, so y = x + 2.

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1 year ago