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3 years ago

How to turn 20/10 into a mixed number

2 answers:

7 0

How you turn 20/10 into a mixed number is you see what the denominator can go into the numerator and in this case it would be 10 and so you see how many times 10 can go into 20 without going over and in this case it would be 2 so the answer would be 2.

Lady_Fox [76]3 years ago

3 0

$\frac{20}{10}$ cannot be changed to a mixed number. 10 could divide both numerator and denominator. And you get 2.

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Tomas is making trail mix using granola and walnuts. He can spend a total of $12 on the ingredients. He buys 3 pounds of granola

Artist 52 [7]

Answer:

Tomas added 6 to both sides of the equation instead of subtracting 6.

Step-by-step explanation:

Tomas is making trail mix using granola and walnuts. He can spend a total of $12 on the ingredients. He buys 3 pounds of granola that costs $2.00 per pound. The walnuts cost $6 per pound. He uses the equation 2x + 6y = 12 to represent the total cost, where x represents the number of pounds of granola and y represents the number of pounds of walnuts. He solves the equation for y, the number of pounds of walnuts he can buy.

Given:

2x + 6y = 12

where

x = number of pounds of granola y = number of pounds of walnuts

The correct solution to the problem

x = 3 pounds

2x + 6y = 12

2(3) + 6y = 12

6 + 6y = 12

Subtract 6 from both sides

6 + 6y - 6 = 12 - 6

6y = 6

Divide both sides by 6

y = 6/6

= 1

y = 1 pound

Tomas added 6 to both sides of the equation instead of subtracting 6.

3 0

3 years ago

Read 2 more answers

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

Radical expression of 4d 3/8

DanielleElmas [232]

Answer:

$\boxed{4 \sqrt[8]{ {d}^{3} } }$

Step-by-step explanation:

$= > 4 {d}^{ \frac{3}{8} } \\ \\ = > 4({d}^{3 \times \frac{1}{8} }) \\ \\ = > 4( {d}^{3} \times {d}^{ \frac{1}{8} } ) \\ \\ = > 4( {d}^{3} \times \sqrt[8]{d} ) \\ \\ = > 4 \sqrt[8]{ {d}^{3} }$