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KATRIN_1 [288]
3 years ago
9

What's the difference between Area and Volume? Explain

Mathematics
2 answers:
Bingel [31]3 years ago
8 0

Answer:

area is how big something is and volume is how much something can hold.

ludmilkaskok [199]3 years ago
5 0

Answer:

An area is length times the width.

A volume is length x width x height

Step-by-step explanation:

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Which of the following is a composite number between 30 and 50? A. 37 B. 33 C. 41 D. 43
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It is answer B I hope I helped u
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O is the centre of the circle, determine the value of v ​
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the angle v =59 degree

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Which ordered pair is the best estimate for the solution to the system
Vera_Pavlovna [14]

Answer:

The best estimate of the solution ordered pair from the graph is (\frac{1}{2},0).

Step-by-step explanation:

See the attached graph to this question.

The graph of two straight lines are shown in the graph.

Now, the two straight lines intersect on the x-axis, so the solution ordered pairs should have y-value equals to zero.

But, there are two ordered pairs with y-value zero and they are (\frac{1}{2},0) and (\frac{1}{3},0).

The best estimate of the solution ordered pairs from the graph is (\frac{1}{2},0).

So, this is the solution. (Answer)

7 0
3 years ago
One angle of an isosceles triangle measures 110°. Which other angles could be in that isosceles triangle? Choose all that apply.
BigorU [14]

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75 an 35 YEE YEE.

Step-by-step explanation:

im pretty sure

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3 years ago
A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11
mixer [17]
Let A(t) denote the amount of salt in the tank at time t. We're given that the tank initially holds A(0)=100 lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}
\implies A'(t)+\dfrac{11}{200+33t}A(t)=484

Find the integrating factor:

\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}

Distribute \mu(t) along both sides of the ODE:

(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}
\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}
A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt
A(t)=22(200+33t)^{2/3}+C

Since A(0)=100, we get

100=22(200)^{2/3}+C\implies C\approx-652.39

so that the particular solution for A(t) is

A(t)=22(200+33t)^{2/3}-652.39

The tank becomes full when the volume of solution in the tank at time t is the same as the total volume of the tank:

200+(44-11)t=500\implies 33t=300\implies t\approx9.09

at which point the amount of salt in the solution would be

A(9.09)\approx733.47\text{ lb}
4 0
3 years ago
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