Let ∠RTS=∠RST = a (say)
∠QUA=∠QSU= b(say)
then we know , at point S, a+40+b=180. so, a+b=140 we'll use this later.
consider trianglePQR, ∠P+∠Q+∠R=180
i.e.P+(180-2b)+(180-2a)=180
P+180+180-2(a+b)=180 ⇒P+180-2(a+b)=0 ⇒P=2(140)- 180=280-180=100
hence,answer is E
Your answer would be
8y^7
-------
3X^3
This is a geometric sequence because each term is twice the value of the previous term. So this is what would be called the common ratio, which in this case is 2. Any geometric sequence can be expressed as:
a(n)=ar^(n-1), a(n)=nth value, a=initial value, r=common ratio, n=term number
In this case we have r=2 and a=1 so
a(n)=2^(n-1) so on the sixth week he will run:
a(6)=2^5=32
He will run 32 blocks by the end of the sixth week.
Now if you wanted to know the total amount he runs in the six weeks, you need the sum of the terms and the sum of a geometric sequence is:
s(n)=a(1-r^n)/(1-r) where the variables have the same values so
s(n)=(1-2^n)/(1-2)
s(n)=2^n-1 so
s(6)=2^6-1
s(6)=64-1
s(6)=63 blocks
So he would run a total of 63 blocks in the six weeks.
It's A), because if the highest exponential would've been uneven, the graph would go up, then down, but as you can see it kinda resembles a parabola, making it one out of A) and B)
as you can see, the graph crosses the system at (0/0), so it can't be C), due to it's constant at the end, being "+1"
so it's A)
