Answer:
H
Step-by-step explanation:
Let represent A and B
A and B is the sum of the first 50 consecutive multiples of 3 and 6, this is the same as a arithmetic sequence because arithmetic sequences have a common sum.
We can represent this as

where a^1 is the first term, d is the common sum, n is the number of multiples we adding.
The first term for both A and B respectively is 3 and 6.
Multiples implies that the Difference between A and B are 3 and 6 respectively. There are 50 consecutively. multiples.
Plug in the known parts for both.


We need to what percent of find 150% of 300.
B is twice as A, so this means that we need to multiply 2 by it original or 100% of it self.

So 200% is the answer.
Answer:
As consequence of the Taylor theorem with integral remainder we have that

If we ask that
has continuous
th derivative we can apply the mean value theorem for integrals. Then, there exists
between
and
such that

Hence,

Thus,

and the Taylor theorem with Lagrange remainder is
.
Step-by-step explanation:
Answer:
-3.8
Step-by-step explanation:
-3.8+7.2=3.4
Answer:
1092
Step-by-step explanation:
Step one:
given data
Ms Lucas has 9 cases of crayon with 52 boxes in each case
total crayon is
9*52= 468
Ms Bruns has 6 cases of crayon with a 104 in each case
total crayon is
6*104= 624
The total crayon is
468+624= 1092
Answer:
a 30 b -21
Step-by-step explanation:
a 5x = 5(6)
= 30
b 3y = 3(-7)
= -21