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3 years ago

15

50 yards of tape is needed to pack and ship 100 packages. You have roll of tape that is 2000 inches long. About how many yards d

o you have?

2 answers:

Olenka [21]3 years ago

5 0

First you need convert inches to yards... one inch is 0.027 of a yard 200 inches is 5.55 yards.

each box requires 0.5 yards of tape with 5.55 yards you can pack 11 boxes with 0.10 of a yard remaining.

castortr0y [4]3 years ago

4 0

56 yards hope this helps

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Robert takes 10 days more than Peter to complete a job. If they work together they will

melamori03 [73]

Answer:

I believe the answer is 2

Step-by-step explanation:

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3 years ago

Kailey buys a greeting card for $3.79. She then buys 3 postcards that all cost the same amount. The total cost is $6.04. How muc

leonid [27]

Answer:

0.75 was how much she paid for each postcard.

Step-by-step explanation:

6.04-3.79

=2.25

so take 2.25 divided by 3

= o.75

equations would be 6.04-3.75=b/ 3

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3 years ago

Select the correct answer.

Pie

Answer:

a.

Step-by-step explanation:

a= y-k/ (x-h)^2

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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

2 years ago

Given: 3x - 5 = 4 (2x – 15)<br> Prove: x = 11<br> Statement<br> Reason

Shalnov [3]

Step-by-step explanation:

3x-5=8x-60

55=5x

x=11

6 0

2 years ago