Idk if this helps but answer that i got was. ) X1=-1,x2=^33/2
Answer:
20-3=17
5-2=3
17/3 is... 5 and 2/3
Step-by-step explanation:
You're right.
Since the output has different number of jumps from one number to another.
Answer: 2(-2x+3)
Step-by-step explanation:
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
