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3 years ago

What is the rate of change

Mathematics

1 answer:

Serggg [28]3 years ago

6 0

Answer:

Slope = 0

Step-by-step explanation:

Let's use the equation $\frac{y2 - y1}{x2 - x1}$ for this problem.

We have two points: (2,0) and (-2, 0)

Let's plug them both into the equation shown above.

$\frac{0 - 0}{-2 - 2}$

Solve.

0 - 0 = 0

-2 - 2 = -4

$\frac{0}{-4}$ = $-\frac{0}{4}$ = 0

A zero in the numerator means this equation has a slope of 0.

Any line with a slope of 0 is a horizontal line.

Slope (rate of change) = 0

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A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per gallon is add

RUDIKE [14]

Salt flows into the tank at a rate of

(1/2 lb/gal) * (6 gal/min) = 3 lb/min

and flows out at a rate of

(Q(t)/60 lb/gal) * (6 gal/min) = 6Q(t) lb/min

The net rate of change of the amount of salt in the tank at time $t$ is then governed by

$\dfrac{\mathrm dQ}{\mathrm dt}=3-6Q$

Solve for $Q$ :

$\dfrac{\mathrm dQ}{\mathrm dt}+6Q=3$

$e^{6t}\dfrac{\mathrm dQ}{\mathrm dt}+6e^{6t}Q=3e^{6t}$

$\dfrac{\mathrm d}{\mathrm dt}[e^{6t}Q]=3e^{6t}$

$e^{6t}Q=\dfrac{e^{6t}}2+C$

$\implies Q=\dfrac12+Ce^{-6t}$

The tank starts with 10 lb of salt, so that Q(0) = 10. This gives us

$10=\dfrac12+C\implies C=\dfrac{19}2$

so that the amount of salt in the tank at time $t$ is given by

$\boxed{Q(t)=\dfrac{1+19e^{-6t}}2}$

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3 years ago

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algol [13]

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2 years ago

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There are a total of 66 students in the

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Answer:

A = 26

Step-by-step explanation:

sum of students = classA + classB + classC

let's say classA = A, classB = B, and classC = C

A + B + C = 66

class A has five more students than class B, so A = 5 more than B so A = 5+B

class C has 2 less students than class B, so C = 2 less than class B = B -2, so C = B-2

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A = 5+B

C = B-2

substitute 5+B for A and B-2 for C in the first equation to limit this to one variable (B)

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1 year ago