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
I believe the answer is 7 since 8x squared is 64x and subtract 4x=60x
60x/16x multiplied by 5/x-5. but i am not sure since i do not know the value of X sorry buddy, just trying to help
Answer:
4
Step-by-step explanation:
She should use 0.74 gallons you have to take 20 and timed it by 3.7 % and thats how you get to see whats left that she need.
Find the line that is normal to the parabola at the given point
remember that normal means perpendicular
perpendicular lines have slopes that multiply to -1
we can use point slope form to write the equation of the line since we are given the point (1,0)
we just need the slope
take derivitive
y'=1-2x
at x=1
y'=1-2(1)
y'=1-2
y'=-1
the slope is -1
the perpendicular of that slope is what number we can multiply to get -1
-1 times what=-1?
what=1
duh
so
point (1,0) and slope 1
y-0=1(x-1)
y=x-1 is da equation
solve for where y=x-1 and y=x-x² intersect
set equatl to each other since equal y
x-1=x-x²
x²-1=0
factor difference of 2 perfect squares
(x-1)(x+1)=0
set to zero
x-1=0
x=1
we got this point already
x+1=0
x=-1
sub back
y=-1-(-1)²
y=-1-(1)
y=-1-1
y=-2
it intersects at (-1,-2)