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ella [17]
3 years ago
12

Simplify the picture below

Mathematics
1 answer:
podryga [215]3 years ago
3 0
((x^2-3x-18)/(x+3))
((x-6)(x+3))/(x+3)
The x+3 on top cancels out the x+3 on the bottom so all you have left is

x-6


Hope I didn't mess up for your sake!
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Can someone help with this?! The answer choices are 1, 1.3, 1.45, and 1.9!
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The answer is 1.45 ......
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The equation y ̂=844.697x^2+3723.485x-13,650 models the annual profits of a new business, where x is the number of years since t
julsineya [31]
The answer is $37,923

<span>y = 844.697x^2 + 3723.485x - 13,650
x - number of years
y - the annual profit

If x = 8, the annual profit y is:
y = </span>844.697 * 8^2 + 3723.485 * 8 - 13,650
y = 844.697 * 64 + 29,787.88 - 13,650
y = 54,060.608 + 16,137.88
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5 0
3 years ago
The is thinking of a number which he calls n. He finds 1/3 of the number and subtracts 5
Alona [7]

Answer:

I just did it its 1/3n - 5

Step-by-step explanation:

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3 years ago
The table shows the decay rate for a certain radioactive material over time. In this context, what does the y-intercept represen
djyliett [7]

Answer:

At 0 hours, there are 100 grams of the substance remaining.

Step-by-step explanation:

4 0
2 years ago
Evaluate the integral by changing to polar coordinatesye^x dA, where R is in the first quadrant enclosed by the circle x^2+y^2=2
34kurt
\displaystyle\iint_Rye^x\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=5}r^2\sin\theta e^{r\cos\theta}\,\mathrm dr\,\mathrm d\theta

which follows from the usual change of coordinates via

\begin{cases}\mathbf x(r,\theta)=r\cos\theta\\\mathbf y(r,\theta)=r\sin\theta\end{cases}

and Jacobian determinant

|\det J|=\left|\begin{vmatrix}\mathbf x_r&\mathbf x_\theta\\\mathbf y_r&\mathbf y_\theta\end{vmatrix}\right|=|r|

Swap the order, so that the integral is

\displaystyle\int_{r=0}^{r=5}\int_{\theta=0}^{\theta=\pi/2}r^2\sin\theta e^{r\cos\theta}\,\mathrm d\theta\,\mathrm dr

and now let \sigma=r\cos\theta, so that \mathrm d\sigma=-r\sin\theta. Now, you have

\displaystyle\int_{r=0}^{r=5}\int_{\sigma=0}^{\sigma=r}re^\sigma\,\mathrm d\sigma\,\mathrm dr=\int_{r=0}^{r=5}r(e^r-1)\,\mathrm dr=4e^5-\dfrac{23}2
7 0
3 years ago
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