For this case we have the following function:

To find the zeros of the function we make
and solve for "x", then:

We multiply by -1 on both sides of the equation:

We factor the equation, for this we look for two numbers that, when multiplied, result in 36 and when added, result in -13. These numbers are -9 and -4.

Thus, the factored equation is:

Therefore, the roots are:

Answer:

Answer:
is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.
Step-by-step explanation:
Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.
First term of given arithmetic progression is A
and common difference is D
ie.,
and common difference=D
The nth term can be written as

pth term of given arithmetic progression is a

qth term of given arithmetic progression is b
and
rth term of given arithmetic progression is c

We have to prove that

Now to prove LHS=RHS
Now take LHS




![=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5BAq%2BpqD-Dq-Ar-prD%2BrD%5D%5Ctimes%20qr%2B%5BAr%2BrqD-Dr-Ap-pqD%2BpD%5D%5Ctimes%20pr%2B%5BAp%2BprD-Dp-Aq-qrD%2BqD%5D%5Ctimes%20pq%7D%7Bpqr%7D)




ie., 
Therefore
ie.,
Hence proved
The annual returns will be calculated as follows:
a] Here we use the formula:
A=p(1+r/100)^n
A=future amount
p=principle
r=returns
n=time
We are given:
A=500, p=400, t=1
Plugging the values in the formula we obtain:
500=400(1+r)^1
simplifying and solving for r:
1.25=1+r
thus
r=1.25-1
r=0.25~25%
b] Using the formula above:
A=p(1+r/100)^n
A=2500+100=2600, p=2000, n=1 year
plugging the values in the equation we obtain:
2600=2000(1+r)^1
simplifying and solving for r we obtain:
2600/2000=1+r
1.3=1+r
hence
r=1.3-1
r=0.3~30%
Answer:
increase
Step-by-step explanation:
11.61 was added to 27