Answer:
<em>B</em> 
Step-by-step explanation:
<u>Dilations</u>
Given a point A(x,y) and a scale factor k the dilated image of A, called A' is calculated as A'=(kx,ky), assuming the same scale factor is applied in both axes.
The pentagon ABCDE was dilated to create pentagon A'B'C'D'E'. To find the dilaton rule used, we must find two clear points where the coordinates of both axes can be easily read from the graph.
Point C(-2,0) maps to C'(-5,0). This gives us the scale factor for the x-axis of -5/(-2)= 5/2.
The y-coordinate of E is 2 and the y-coordinate of E' is 5. This gives us the same scale factor for the y-axis of 5/2.
Thus, the rule to dilate the pentagon is:
B 
Since the divisor is in the form (x + #) or (x - #), This can be done by synthetic division.
First put the polynomial ion descending order: x^2 - 7x + 15
Take the coefficients of the terms and follow these steps:
3 | 1 -7 15
3 -12
___________ Bring down the 1, multiply the 3 by the 1 and place under the
1 -4 3 -7, then add.
Multiply 3 by -4, place under the 15, then add.
The bottom row is our answer. Since the problem started with a second power, the answer will start with a first power.
The bottom row are the coefficients of the terms and the last number is the remainder.
x - 4 remainder 3 ALSO WRITTEN x - 4 + 3/(x -3)
40,023,032 = (4 x 1000000000) + (0 x 100000000) + (0 x 10000000) + (0 x 1000000) + (2 x 100000) + (3 x 10000) + (0 x 1000) + (0 x 100) + (3 x 10) + (2 x 1)
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up
going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r
notice, the distance is the same, upstream as well as downstream
thus
![\bf \begin{cases} b=\textit{rate of the boat}\\ r=\textit{rate of the river} \end{cases}\qquad thus \\\\\\ \begin{array}{lccclll} &distance&rate&time(hrs)\\ &----&----&----\\ upstream&48&b-r&4\\ downstream&48&b+4&3 \end{array} \\\\\\ \begin{cases} 48=(b-r)(4)\to 48=4b-4r\\\\ \frac{48-4b}{-4}=r\\ --------------\\ 48=(b+r)(3)\\ -----------------------------\\\\ thus\\\\ 48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ab%3D%5Ctextit%7Brate%20of%20the%20boat%7D%5C%5C%0Ar%3D%5Ctextit%7Brate%20of%20the%20river%7D%0A%5Cend%7Bcases%7D%5Cqquad%20thus%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26distance%26rate%26time%28hrs%29%5C%5C%0A%26----%26----%26----%5C%5C%0Aupstream%2648%26b-r%264%5C%5C%0Adownstream%2648%26b%2B4%263%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0A%0A%5Cbegin%7Bcases%7D%0A48%3D%28b-r%29%284%29%5Cto%2048%3D4b-4r%5C%5C%5C%5C%0A%5Cfrac%7B48-4b%7D%7B-4%7D%3Dr%5C%5C%0A--------------%5C%5C%0A48%3D%28b%2Br%29%283%29%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Athus%5C%5C%5C%5C%0A48%3D%5Cleft%5B%20b%2B%5Cleft%28%5Cboxed%7B%5Cfrac%7B48-4b%7D%7B-4%7D%7D%5Cright%29%20%5Cright%5D%20%283%29%0A%5Cend%7Bcases%7D)
solve for "r", to see what the stream's rate is
what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
That creates the number 82,190
