By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is
<h3>What is
sequence ?</h3>
Sequence is collection of numbers with some pattern .
Given sequence

We can see that

and

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2
Now
term of this Geometric progression can be written as

So summation of 15 terms can be written as

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is
To learn more about Geometric progression visit : brainly.com/question/14320920
Answer:
the common differnence is positive 3
Step-by-step explanation:
its going back by 3 since its going back to 0 its positive
Answer:

Step-by-step explanation:
we are given surface area and the length of the square base
we want to figure out the Volume
to do so
we need to figure out slant length first
recall the formula of surface area

where B stands for Base area
and P for Base Parimeter
so

now we need our algebraic skills to figure out s
simplify parentheses:

reduce fraction:

simplify multiplication:

cancel 64 from both sides;

divide both sides by 16:

now we'll use Pythagoras theorem to figure out height
according to the theorem

substitute the value of l and s:

simplify parentheses:

simplify squares:

cancel 16 from both sides:

square root both sides:

recall the formula of a square pyramid

where A stands for Base area (l²)
substitute the value of h and l:

simplify multiplication:

reduce fraction:

hence,

Answer:
about 20
Step-by-step explanation:
its kindergarden math
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C




![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)


LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 