1/2 + 3/4 + 7/8
4/4(1/2)+ 2/2(3/4) + 7/8
4/8 + 6/8 + 7/8
4+ 6 + 7
—————
8
17 1
——. Or 2—-
8 8
He prepared 2 1/8 lbs of trail mix
G(f(x)) = g(2x +7) = -3(2x +7) +9 = -6x -21 +9
g(f(x)) = -6x -12
QUESTION:
The code for a lock consists of 5 digits (0-9). The last number cannot be 0 or 1. How many different codes are possible.
ANSWER:
Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:–
- P(n,r) = n!/(n−r)! where n is the number of numbers in the set and r is the subset.
Since there are 10 digits to choose from, we can assume that n = 10.
Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r = 5.
Now, plug these values into the formula and solve:
= 10!(10−5)!
= 10!5!
= 10⋅9⋅8⋅7⋅6
= 30240.
A relation is <em>not</em> a function if it has repeated "x" values.
A. (3, _) repeats
B. is a function
C. (5, _) repeats
D. (-4, _) repeats
Answer:
Step-by-step explanation:
Evaluate 6(5-(6-5)-3) = 6(5-1-3) {first solve innermost bracket}
= 6* (5-4) = 6*1 =6
