Answer:
Yes its similar because if you look at all the numbers they are all the same but the shape is just turned around
please consider brainliest <3
8z-7=81
8z=88
z=11
Have a good day!
:0)
Answer:
BC = 24
Step-by-step explanation:
FD = 36, FE = 15 and CD = 18
By secant theorem.
DE • DF = CD • DB
(36 - 15) • 36 = 18 • DB
DB = 42
BC = DB - CD
BC = 42 - 18
BC = 24
PLEASE MARK ME AS BRAINLIEST AND HAVE A NICE DAY :)
Answer:
42 * 10 + 42 * 5
which agrees with the third option (C) shown on the list of possible answers
Step-by-step explanation:
Notice that using distributive property we can write:
42 * 15 = 42 (10 + 5) = 42 * 10 + 42 * 5
which agrees with the third option (C) shown on the list of possible answers.
Option C:
The coefficient of
is 40.
Solution:
Given expression:

Using binomial theorem:

Here 
Substitute in the binomial formula, we get

Now to expand the summation, substitute i = 0, 1, 2, 3, 4 and 5.


Let us solve the term one by one.






Substitute these into the above expansion.

The coefficient of
is 40.
Option C is the correct answer.