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balu736 [363]
3 years ago
14

Use the graph to answer F(-3)=

Mathematics
2 answers:
Virty [35]3 years ago
7 0

Answer:

  f(-3) = 1

Step-by-step explanation:

The only point with an x-coordinate of -3 has a y-coordinate of 1. That is the value of the function f(x) for x=-3.

  f(-3) = 1

Lynna [10]3 years ago
7 0
There is only one point on -3.
So the answer is
f(-3)=1
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Ok so this is so confusing The volume of water in a rectangular swimming pool can be modeled by the polynomial 2x3 _ 9x2 + 7x +
Alika [10]
The volume formula is V= l x L x H, l=width, L=Length, H= Depth, so 
2x3 _ 9x2 + 7x + 6 = l x L x (2x + 1), because H=(2x + 1), so 
l x L= (2x3 _ 9x2 + 7x + 6 )/ (2x + 1) = (2x3 _ 9x2 + 7x + 6 ) X [1/(2x + 1)]
case1: l= (2x3 _ 9x2 + 7x + 6 ) or L= 1/(2x + 1), case2: L= (2x3 _ 9x2 + 7x + 6 ) or l= 1/(2x + 1)
the why question:
perhaps there is similarity of value between volume and l, or volume and L
6 0
3 years ago
You want to rent a limousine for a trip to the city. The limo costs $700 for the night and $0.15 per mile. You have $750 to spen
mr Goodwill [35]
700+0.15m=750

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5 0
3 years ago
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The national collegiate athletic association (ncaa) uses a sliding scale for eligibility for division i athletes. those students
Blizzard [7]

A suitable probability calculator pegs that probability at 81%.

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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
Hey can anyone pls answer these math problems!!
Zepler [3.9K]

Answer:

1={5 and 3} 2={0} 3={0.209 and 4.791} 4={-3 and 4} 5={1 and 9} 6={-4.303 and -0.697}

Step-by-step explanation:

Your probably confused on what a, b, and c means.

The problems are written like this.

ax^2+bx+c=0

That formula finds the zeros.

For example 1:

x^2-8x+15=0\\ax^2-bx-c\\a=1\\b=-8\\c=15

Then use the formula

x=\frac{-b+-\sqrt{b^2-4ac} }{2a} \\\\x=\frac{-(-8)+-\sqrt{-8^2-4(1)(15)} }{2} \\x=\frac{8+-\sqrt{64-60} }{2} \\\\x=\frac{8+-2}{2}\\

If plus then 5, if minus then 3

CHECK:

x^2-8x+15=0\\5^2-8(5)+15=0\\25-40+15=0\\0=0\\3^2-8(3)+15=0\\9-24+15=0\\0=0

Hope this helps you understand how to do this by yourself.  Now to do the rest ;-;

5 0
2 years ago
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