The answer is B! b^4. If we examine the expression given, we have a^3 * b^3 * a^-3 * b= a^3/a^3 * b^3 * b = 1 * b^4
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The answer is 5
Here are the steps:
First off, we will be using the distance formula of
So we have the ordered pairs of (3,1) and (6,5)
Once you plug them into the formula it should look like this:
Now we do the math inside the parenthesis and end up with:
Then you multiply by the power and simplify to get:
And the
=5
So your answer is
5
Answer:
first we find the common difference.....do this by subtracting the first term from the second term. (9 - 3 = 6)...so basically, ur adding 6 to every number to find the next number.
we will be using 2 formulas....first, we need to find the 34th term (because we need this term for the sum formula)
an = a1 + (n-1) * d
n = the term we want to find = 34
a1 = first term = 3
d = common difference = 6
now we sub
a34 = 3 + (34-1) * 6
a34 = 3 + (33 * 6)
a34 = 3 + 198
a34 = 201
now we use the sum formula
Sn = (n (a1 + an)) / 2
S34 = (34(3 + 201))/2
s34 = (34(204)) / 2
s34 = 6936/2
s34 = 3468 <=== the sum of the first 34 terms:
Answer:
60 yards back
Step-by-step explanation:Mutiply 4 and 15
<span>since sin and cos = each other at pi/4; take your integrals from 0 to pi/4
</span><span>[S] cos(t) dt - [S] sin(t) dt ;[0,pi/4]
</span>
<span>to revolve it around the x axis;
we do a sum of areas
[S] 2pi [f(x)]^2 dx
</span>
<span>take the cos first and subtract out the sin next; like cutting a hole out of a donuts.
</span><span>pi [S] cos(x)^2 dx - [S] sin(x)^2 dx ; [0,pi/4]
</span>
<span>cos(2t) = 2cos^2 - 1
cos^2 = (1+cos(2t))/2
</span>
<span>1/sqrt(2) - (-1/sqrt(2) +1)
1/sqrt(2) + 1/sqrt(2) -1
(2sqrt(2) - sqrt(2))/sqrt(2) = sqrt(2)/sqrt(2) = 1</span>
