By polynomial grid division, we start by the divisor x²-3x+4 placed on the row headings of the table and end with the quotient -2x + 5 on the column headings as given. We know that -2x³ must be in the top left which means that the first column entry is indeed -2x. So the row and column multiply to -2x³. We use this to fill in all of the first column, multiplying -2x by the terms of the row entries.
-2x 5
x² -2x³
-3x 6x²
4 -8x
We now got 6x² though we want 11x². The next quadratic entry must then be 5x² so that the overall sum is 11x². Multiplying 5 by the terms of the row entries, we fill in all of the second column:
-2x 5
x² -2x³ 5x²
-3x 6x² -15x
4 -8x 20
The bottom and final term is 20, which is our desired answer and we can read the quotient off the first row:
-2x³ +11x² - 23x + 20 / x² - 3x + 4 = -2x + 5
We have calculated for all the terms that belong in table, therefore, the terms 5x² and 6x² belong in the shaded cells.
Answer:
y=x-5
Step-by-step explanation:
sale price (y) is equal to the original (x) minus five dollars
I’m thinking it’s $100. I guessed 100 first as that is a pretty common number and subtracted 61.75. I was left with 38.25 and I divided that by 3, the amount of washes. I got $12.75 but I had to check since I guessed 100. I then subtracted 23.50 from 100 and got 76.5. I divided that by 6, the amount of washes, and got 12.75 again. So I’m pretty sure that there was $100 on the gift card before he washes his car six times, each wash being $12.75.
Answer:
35.00% increase
Explanation:
[(New value - Old value)/Old value] x 100 = answer
Hope this helps in the future
Suppose order does not matter, then combinations will consist of 0 to 3 of the singleton colors, and 5 to 8 that are some combination of red, white or blue. This implies that we shall formulate the answer as the sum from k=0 to k=3 of:
N=(Number of ways to pick k of the single colors)*(number of ways to pick 8-k balls from red, white pr blue)
The first factor is C(3,k), let's call the second factor A(8-k), where A(n) is the number of combinations or red, white or blue when n balls are selected. Given that we want to express A(n) as a formula in terms of n, let's say there are r red balls in a combination (0≤r≤n). Then there are (n-r) balls that are either white or blue that have been left over. Thus, the number of white balls must be integer from 0 to (n-r), for the universe of (n-r+1) possibilities. Hence, the rest are blue in exactly one way, so A(n) is the sum over r=0 to n:
A(n)=∑(1+n-r)
A(n)=(n+1)(n+1)-n(n+1)2
=(n+1)(2n+2-n)/2
=(n+1)(n+2)/2
thus
A(5)=6*7/2=21
A(6)=7*8/2=28
A(7)=8*9/2=36
A(8)=9*10/2=45
The total will give:
N=C(3,0)*A(8)+C(3,1)*A(7)+C(3,2)*A(6)+C(3,3)*A(5)
N=1*45+3*36+3*28+1*21=258
The answer is 258