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3 years ago

HELP ASAP!! What is the average rate of change between x = -3 and x = 4 for the equation y = 8x2?

Mathematics

1 answer:

Llana [10]3 years ago

3 0

Answer:

$x=-3\\y=8(-3)^2\\y=8(9)\\y=72$

$x=4\\y=8(4)^2\\y=8(16)\\y=128$

128-72=56

average rate of change is 56

Step-by-step explanation:

Plz let me know if im right

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If (6^2)^x equals 1, what is the value of x

ValentinkaMS [17]

Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

At his favorite restaurant, Dylan orders a soda with his dinner. The ice in his drink has a mass of 6 g and a volume of 20 cm3.

frez [133]

Answer:

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Step-by-step explanation

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m=mass

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p=density

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<u>Hope this helped!</u>

5 0

2 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

Stells [14]

$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

3 0

3 years ago

I found a place that will give me a 20 % discount if i spend over $50

anygoal [31]

You save $15

Work:

75 x 0.20 (20%) = 15

4 0

3 years ago