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1+2=2

2+4=4

4+4=8

8+8=16

1x1=1

1x2=2

2x2=4

4x4=8

8x8=64

1,2,3,4,5,6,7,8,9,0

h(x)=5x-1 find: h9a-1)

h(a-1)=5(a-1)1

=a-5-1

=5a-6

3x-2y>6

3(0)-2(0)>6

0-0>6

0>6 false

Distance varies directly as mass.

Distance is 20 when mass is 3.

Find distance when mass is 5.

The way these equations are presented lends itself to the substitution method (or Y)

What I'm going to do:

Substitute 2 for X

in the formula/simplify

f(x)=-3x^2+6x+2

-3(x^2-2x+1)+3+2

-3(x-1)^2+5

-7i(3+4i)

=-21i+(-28i)

=21i+(-28i^2)

=-21i+(-28(^-1))

=28-21i