Weird way to write it but alright! (Sideways)
19pq^-2 x 5pq^6 = ?
These problems are pretty much single operations between each of the variables / constants.
So it's like this:
(19*5)(p*p)(q^-2*q^6) = ?
19*5 is 95.
For p*p remember that when two variables multiply there given powers add. In the case where the powers are not shown (like in the case of p*p) they are always assumed to be 1. So what is 1+1? 2.
p*p is p^2
For q^-2*q^6 it is the same deal with the previous problem. So now the problem looks like this:
-2 + 6 = 4
(The two is negative, because the power is negative 2)
So, q^4.
Our final answer is all of the combined.... like a so:
95p^2q^4
Answer:
x)45
y)45
z)135
Step-by-step explanation:
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: The answer is D, 5/6.
Step-by-step explanation: The reciprocal of a fraction is that fraction but the numerator and denominater swapped places.
