Answer:
see below
Step-by-step explanation:
(ab)^n=a^n * b^n
We need to show that it is true for n=1
assuming that it is true for n = k;
(ab)^n=a^n * b^n
( ab) ^1 = a^1 * b^1
ab = a * b
ab = ab
Then we need to show that it is true for n = ( k+1)
or (ab)^(k+1)=a^( k+1) * b^( k+1)
Starting with
(ab)^k=a^k * b^k given
Multiply each side by ab
ab * (ab)^k= ab *a^k * b^k
( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)
Therefore, the rule is true for every natural number n
Answer: 
Step-by-step explanation:
Answer:
<u>Infotron should produce each day 13 hockey and 6 soccer games</u>
Step-by-step explanation:
x = Number of hockey games Infotron should produce
y = Number of soccer games Infotron should produce
Number of labor-hours for assembly = 2x + 3y
Number of labor-hours for testing = 2x + y
Now we can write our equations system, this way:
2x + 3y = 44
2x + y = 32
*********************
Expressing y in terms of x in the 2nd equation:
2x + y = 32
y = 32 - 2x
********************
Substituting y and solving for x in the 1st equation:
2x + 3y = 44
2x + 3 * ( 32 - 2x) = 44
2x + 96 - 6x = 44
-4x = 44 - 96
-4x = - 52
x = -52/-4
x = 13
*****************
Solving for y in the 2nd equation:
2x + y = 32
2 * 13 + y = 32
26 + y = 32
y = 32 - 26
y =<u> 6</u>
<u>Infotron should produce each day 13 hockey and 6 soccer games</u>
(2/3)x^2 -6x + 15 = 0
Using the quadratic formula:
x = [-b +-sq root(b^2 - 4 *a*c)] / 2a
x= [--6 +-sq root(36 -4*(2/3)*15] / 2*(2/3)
x= [6 +-sq root 36 -40] / (4/3)
x1 = 4.5 + (2i / (4/3))
x1 = 4.5 + 1.5i
x2 = 4.5 - (2i / (4/3))
x2 = 4.5 - 1.5i
64000 x 0.12 = 7680
she was given a $7,680 bonus
