The width of the driveway is 5x-2.
The height of the carport is 6x+1.
We complete polynomial long division for both of these.
For the driveway:
Dividing 5x²+43x-18 by x+9, we first see how many times x will go into 5x². It goes in 5x times; we write this in the division problem above 43x. Multiplying back through,
5x(x+9) = 5x²+45x. This goes below 5x²+43x. We now subtract:
5x²+43x-(5x²+45x)= -2x; this goes below the 45x we wrote earlier. Bring down the -18.
Now we see how many times x goes into -2x; it goes -2 times. This goes beside our 5x in our answer at the top. Multiplying back through,
-2(x+9) = -2x-18. This goes below our -2x-18 we had, and gives us an answer of 5x-2 with a remainder of 0.
For the carport:
To divide 48x³+68x²-8x-3 by 8x²+10x-3, we see how many times 8x² will go into 48x³. It will go 6x times; write this above -8x. Multiplying back through,
6x(8x²+10x-3) = 48x³+60x²-18; write this below 48x³+68x²-8x. Now subtract:
48x³+68x²-8x-(48x³+60x²-18) = 8x²+10x; this goes below our 48x³+60x²-18. Bring down the -3.
Now we want to see how many times 8x² will go into 8x². It goes 1 time; write this beside our 6x at the top. Multiplying back through,
1(8x²+10x-3) = 8x²+10x-3; write this below the 8x²+10x-3 we have already down. When we subtract these, we get a remainder of 0, with our answer up top as 6x+1.
6 to 10 10-6 = 4 that is the range
There are two possible outcomes of this experiment either success p or failure q. It has a given number of trials and all trials are independent therefore it is<u><em> binomial probability distribution.</em></u>
1- 5 ways
2- 5/16
3- 1/16
4- 1/16
In the question given above n= 5 p =1/2 q= 1/2 r is the given point.
- <u>Part 1:</u>
The number of ways in which different people get off the bus can be calculated using combinations since the order is not essential. Therefore
nCr= 5C4= 5 ways
<u>2. Part 2:</u>
The probability that all four people get off the bus on the first stop is given by :
P (x= 1)= 5C1 (1/2)^0(1/2)^4= 5(1/2)^4= 5/16
<u>3. Part 3:-</u> The probability that all four people get off the bus on the same stop.
P (x= x)= 5C5 (1/2)^0(1/2)^4= 1(1/2)^4= 1/16
<u>4. Part 4-</u> The probability that <u><em>exactly three of the four</em></u> people get off the bus on the same stop.
P (x= x)= 5C5 (1/2)^3(1/2)^1= 1(1/2)^4= 1/16
For binomial distribution click
brainly.com/question/15246027
brainly.com/question/13542338
F(x) = 3x+1
G(x) = X^2 - 6
F(G(x)) = F(X^2 - 6) = 3(X^2 - 6) + 1 = 3X^2 - 18 + 1 = 3X^2 -17
F(G(x)) = 3X^2 - 17
3y+6 is the answer
-9+5(y+3)-2y
-9+5y+15-2y
-9+3y+15
=3y+6
