Answer:
$6.03
Step-by-step explanation:
The question is asking us how much each pair of socks cost. With the information we are given, we know that there are 5 pairs and that they all cost the same amount. If the total were just the 5 socks, we could easily divide the total by 5 to get the price of each pair but that's not the case. The soccer ball is included in the final cost. To overcome this, we just have to subtract the soccer ball's cost ($45) from the total.
So we subtract $45 from $75.15, giving us $30.15, the total amount of all the socks. However we need to find out how much each pair costs and there are 5 of them, so we divide 30.15 by 5. This gives us the answer, $6.03.
Each pair of socks cost $6.03.
Simpler Explanation:
- $75.15 - $45 = $30.15
- $30.15 / 5 = $6.03
The answer is $6.03
First, we must find the common denominator (12) ✵
10/12-6/12 ✵ ✵ ✵ ✵
Now, why 10? ✵ ✵ ✵
Because we CAN'T just take the denominator and change it; we must change both the denominator AND the numerator:
✵ • - ○
4/12 ✧ ✩ ✶ ✺ ✱
We can simplify, or reduce, this fraction: ✱ ✱
1/3 ✩ ✦ ✤
I hope it helps!
Answer:
h(8q²-2q) = 56q² -10q
k(2q²+3q) = 16q² +31q
Step-by-step explanation:
1. Replace x in the function definition with the function's argument, then simplify.
h(x) = 7x +4q
h(8q² -2q) = 7(8q² -2q) +4q = 56q² -14q +4q = 56q² -10q
__
2. Same as the first problem.
k(x) = 8x +7q
k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q
_____
Comment on the problem
In each case, the function definition says the function is not a function of q; it is only a function of x. It is h(x), not h(x, q). Thus the "q" in the function definition should be considered to be a literal not to be affected by any value x may have. It could be considered another way to write z, for example. In that case, the function would evaluate to ...
h(8q² -2q) = 56q² -14q +4z
and replacing q with some value (say, 2) would give 196+4z, a value that still has z as a separate entity.
In short, I believe the offered answers are misleading with respect to how you would treat function definitions in the real world.
Answer:
Yes.
Step-by-step explanation:
Just like normal algebra, you factor our the common factor, in this case, 5.
Thus,
