May be there is an operator missing in the first function, h(x). I will solve this in two ways, 1) as if the h(x) = 5x and 2) as if h(x) = 5 + x
1) If h(x) = 5x and k(x) = 1/x
Then (k o h) (x) = k ( h(x) ) = k(5x) = 1/(5x)
2) If h(x) = 5 + x and k (x) = 1/x
Then (k o h)(x) =k ( h(x) ) = k (5+x) = 1 / [5 + x]
Answer:
The 1st error that Carlos made was in Step 2...
In step 2 , Carlos multiplied 14 only with 40. If he would have multiplied 14 with both 40 & 9 , then it could have been alright.....
Answer: 
Step-by-step explanation:
<h3>
The exercise is: " Evaluate 
when 
and 
</h3>
Given the following expression:
You can follow these steps in order to evaluate it:
1. Substitute and into the expression provided in the exercise:
2. Solve the multiplications. Remember that:
Then:
3. Reduce the fraction. Notice that the numerator 6 and the denomiantor 4 can be both divided by 2. Then:
4. Solve the addition:
Since the number 21 has a denominator 1, the Least Common Denominator is:
Then, the sum is:
Answer:
so you can make both your dividend and your divisor equal so you can divide
Step-by-step explanation:
Of the 27 players trying out for the school basketball team, 8 are more than 6 feet tall and 7 have good aim. What is the probability that the coach would randomly pick a player over 6 feet tall or a player with a good aim? Assume that no players over 6 feet tall have good aim. A. 7/15 B. 6/15 C. 7/9 D. 5/9
The Answer Is
--------5/9--------
