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1 / 2*9*20*4 + 9 squared, is 1 / 1440 + 81. This, if it is what I think it is, is 1 / 1521 because you never said whether or not there were perenthesis around the 2*9*20*4 + 9. Otherwise, the answer would be 123201/1521, or 81. 

For part c, after you draw the two segmants from PA and PF, we can see that triangle FPE can fit on GF, while triangle ADP can fit on AB. This creates a third square that is diagonal. 

looks like we got a homework cheater on our hands!

\(\displaystyle (1+x)(a-bx)^{12} = \\ \displaystyle (1+x)\{(^{12}_{\, 0})a^{12}+( ^{12}_{\, 1})a^{11}(-bx)+( ^{12}_{ \,2})a^{10}(-bx)^{2}+( ^{12}_{ \, 3})a^{9}(-bx)^{3}+\dots \} \\ \displaystyle = (1+x)(a^{12}-12a^{11}bx+66a^{10}b^{2}x^{2}-220a^{9}b^{3}x^{3}+ \dots) \\ \displaystyle = a^{12}+(a^{12}-12a^{11}b)x-(12a^{11}b-66a^{10}b^{2})x^{2}+(66a^{10}b^{2}-220a^{9}b^{3})x^{3}+\dots \)

If the coefficient of the x-cubed term is zero, then

\(\displaystyle 66a^{10}b^{2}=220a^{9}b^{3} \\ \displaystyle \frac{a}{b}=\frac{220}{66}=\frac{10}{3}.\)

Thank you for answering, you saved my day :)

Your question is incorrect I think, I might be wrong, but r can be both 1 and 9. 

To solve this problem, we can set up an equation based on the given information.

Let's assume the distance you drove is represented by 'd' miles.

When driving at a rate of 40 miles per hour, the time it took is given by:

Time1 = d / 40

When driving at a rate of 25 miles per hour, the time it took is given by:

Time2 = d / 25

According to the problem, if you had driven at 25 miles per hour instead, you would have arrived 45 minutes (3/4 of an hour) later. So we can set up the equation:

Time2 = Time1 + 3/4

The answer is 4 because in 1 step, you can only go to (0,1), (0, -1), (1, 0), (-1, 0). 

Sorry, I think I read the question wrong. The answer was not 13 :(

the answer will be 5/4^98

I think the answer is 62 ways.

The answer is 13. 
I don't have time right now to tell you how I did it. 
Sorry it. 

Very true. Never considered that. 

By me, GA!

that's because they are a homework cheater, and they don't know!

Isn't this answer 12i + 8j? You never specified what the variables were, or whether I should have variables in the answer. 

We can suggest that the distance in miles was d. 

d / 40 is time at 40mph

d / 25 is time at 25mph

Through this, we can see that d / 40 + 45 / 60 = d / 25

45 / 60 is 3 / 4, so we get d / 40 + 3 / 4 = d / 25. 

We can get the LCM(Least Common Multiple) of 40, 4, 25, which is 200, and we multiply both sides of d / 40 + 3 / 4 = d / 25 by 200. 
 

This now makes 5d + 150 = 8d. We can now subtract 5d from both sides, which is 3d= 150, then divide both sides by 3, and get d = 50. 

I think this question has been already asked, the link is:

https://web2.0calc.com/questions/need-help-with-system_6 

We first try to find the length of PM, so we divide PQ by 2. PM = PQ/2 = 36/2 = 18

Then, we use the bisector theorem to find the lengths of PX and QX. Because PX bisects the angle QPR, PX / RX = PQ / RQ

Through subsitution, we can get 36/26, or 18/13. 

PX = (18 * 18) / 13 = 24.92, and RX = (13 * 18) / 13 = 18

Using the Pythagorean theorem, we can find AX. AX² = PX² + RX²       AX² = 24.92² + 18²                AX² = 621 +324        AX = sqrt945       AX = 30.74. 

Because Y is on the perpendicular bisector of PQ, PY = QY = PQ / 2 = 18

Therefore, AY = AX - XY = 30.74 - 8, or 22.74

Finally, we can use Heron's formula to find the area of the triangle PQR. s = (36 + 22 + 26) / 2 = 42

area(PQR) = sqrt(s(s-36)(s-22)(s-26)) = sqrt(42*6*20*16) = 336

Therefore, the area of triangle PQR is 336 square units.