Group and factor
undistribute then undistribute again
remember
ab+ac=a(b+c)
this is important
6d^4+4d^3-6d^2-4d
undistribute 2d
2d(3d^3+2d^2-3d-2)
group insides
2d[(3d^3+2d^2)+(-3d-2)]
undistribute
2d[(d^2)(3d+2)+(-1)(3d+2)]
undistribute the (3d+2) part
(2d)(d^2-1)(3d+2)
factor that difference of 2 perfect squares
(2d)(d-1)(d+1)(3d+2)
77.
group
(45z^3+20z^2)+(9z+4)
factor
(5z^2)(9z+4)+(1)(9z+4)
undistribuet (9z+4)
(5z^2+1)(9z+4)
The correct answer would be rectangle 'D'.
If we looked at rectangle 'E' and moved it to the left by 7 units (every unit = 1 square on the grid), we would arrive at a point. From then, we can move upwards 12 units and arrive at rectangle 'D'.
Proof by induction
Base case:
n=1: 1*2*3=6 is obviously divisible by six.
Assumption: For every n>1 n(n+1)(n+2) is divisible by 6.
For n+1:
(n+1)(n+2)(n+3)=
(n(n+1)(n+2)+3(n+1)(n+2))
We have assumed that n(n+1)(n+2) is divisble by 6.
We now only need to prove that 3(n+1)(n+2) is divisible by 6.
If 3(n+1)(n+2) is divisible by 6, then (n+1)(n+2) must be divisible by 2.
The "cool" part about this proof.
Since n is a natural number greater than 1 we can say the following:
If n is an odd number, then n+1 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.
If n is an even number" then n+2 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.
Therefore by using the method of mathematical induction we proved that for every natural number n, n(n+1)(n+2) is divisible by 6. QED.
Answer:
neither
perpendicular
parallel
im pretty sure
Step-by-step explanation:
Step-by-step explanation:
7. Since they are similar triangles we have:
So DF is 15
8. They are also similar. So:
So PR is 10
