Find a number less than 100 that is increased by one-fifth of its value when its digits are reversed.
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
In this question, the sample space contains 8 elements and has been given as;
HHT HHH THH HTH HTT TTT TTH THT
1. For event A:
Outcomes of alternating tail and head = THH HTH HTT THT
= 4 outcomes
Pr(alternating tail and head (with either coming first)) =
=
2. For event B:
Outcomes of no tails on the first two tosses = HHT HHH
= 2 outcomes
Pr (No tails on the first two tosses) =
=
For event C:
Outcomes of a tail on both the first and the last tosses = THT TTT
= 2 outcomes
Pr(A tail on both the first and last tosses) =
=
Answer:
4x⁻¹
Step-by-step explanation:
4x + 20 25 - x²
_______ ÷ _______ ↷
x² - 5x [x - 5]²
4[x + 5] [-5 + x][5 + x]
_______ ÷ _______
x[x - 5] [x - 5][x - 5]
So what we did here was factor out ALL the Greatest Common Factors [GCFs] in each denominator and numerator, and since we are dividing vertically AND horizontally, according to this, ALL [x - 5] and [x + 5] cancel each other out and all you are left with is this:
4x⁻¹
OR
This, according to the Negative Exponential Rule [Reverse], is how you rewrite it as a fraction. You bring the denominator to the numerator while ALTERING THE INTEGER SYMBOL FROM POSITIVE TO NEGATIVE:
b⁻ⁿ = 1\bⁿ
I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.