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alekssr [168]
3 years ago
6

4xy - 6xy + 5y - 9y Please also include the steps to solving it.

Mathematics
1 answer:
gizmo_the_mogwai [7]3 years ago
8 0

Answer:

-2xy -4y

Step-by-step explanation:

Combine like terms:

Subtract 6xy from 4xy: 4xy-6y = -2xy

Combine Like terms:

Subtract 9y from 5y: 5y-9y= -4y

Ans: -2xy -4y

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Help calculus module 8 DBQ<br><br> please show work
igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of R(t) at the endpoints of each subinterval. The sum is then

\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since R is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval [a,b], it guarantees the existence of some c\in(a,b) such that

\dfrac{R(b)-R(a)}{b-a)=R'(c)

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such c for which R'(c)=0. I would chalk this up to not having enough information.

3. R(t) gives the rate of water flow, and R(t)\approx W(t), so that the average rate of water flow over [0, 8] is the average value of W(t), given by the integral

R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt

If doing this by hand, you can integrate by parts, setting

u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt

\mathrm dv=\mathrm dt\implies v=t

R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)

For the remaining integral, consider the trigonometric substitution t=\sqrt 7\tan s, so that \mathrm dt=\sqrt 7\sec^2s\,\mathrm ds. Then

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)

\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}

or approximately 3.0904, measured in gallons per hour (because this is the average value of R).

4. By the fundamental theorem of calculus,

g'(x)=f(x)

and g(x) is increasing whenever g'(x)=f(x)>0. This happens over the interval (-2, 3), since f(x)=3 on [-2, 0), and -x+3>0 on [0, 3).

5. First, by additivity of the definite integral,

\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt

Over the interval [-2, 0), we have f(x)=3, and over the interval [0, 6], f(x)=-x+3. So the integral above is

\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}

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Answer is option C.

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juin [17]
In math "of" means times, so...

2/3 of 800 =
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