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3 years ago

A student said, “To find the value of 109.2÷6, I can divide 1,092 by 60.”

Mathematics

1 answer:

Juliette [100K]3 years ago

4 0

Answer:

I agree with the statement.

Step-by-step explanation:

As long as both numbers are being multiplied by the same quantity, you will always end up with the same number.

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Factorize<br>3a⁴+18a²-21-2b-3b²

Gnom [1K]

Answer:

the expression is not factorable with rational numbers.

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3 years ago

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What set of reflections would carry parallelogram ABCD onto itself?

My name is Ann [436]

"y-axis, x-axis, y-axis, x-axis" is the set of reflections among the following choices given in the question that <span>would carry parallelogram ABCD onto itself. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that has helped you.</span>

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3 years ago

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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

I need help I’m stuck

cluponka [151]

Answer:

Step-by-step explanation:

5 0

3 years ago

A recipe calls for 5 deciliters of water and 236 milliliters of milk. How many milliliters of liquid does the recipe call for in

olya-2409 [2.1K]

1 deciliter is one-tenth of a liter and 1 milliliter is one-thousandth of a liter.
So: 1 liter = 10 deciliter = 100 centiliter = 1000 milliliter

This means that 1 deciliter is equal to 100 milliliters.
Based on the above:

5 deciliters = 500 milliliters
500+236 = 736

The recipe calls for 736 milliliters of liquid in total.

3 0

3 years ago