Answer:
43
Step-by-step explanation:
g(x)=4x+3
plug in 10
g(10)= 4(10)+3
= 40+3
= 43
This is a common factor problem.
Pencils come in a pack of 12
Erasers come in a pack of 10
First, break the number into their prime factors(the idea is that we will break the number down into its smallest multiples, which are prime numbers):
10 = 2 * 5
12 = 2 * 2 *3
So now we take the unique multiples of each number, and when we multiply them together, we will get the smallest number that both 10 and 12 can be divided into(this is what the problem is asking for)
We have (2*2*3) that comes from 12, and the only unique number that comes from the 10 is (5)
So now, we multiply:
2*2*3*5=60
However, this isn't exactly out answer. Now we have to divide our answer by the number of each this in the pack to know how many packs to buy.
60/12=5 packs of pencils
60/10= 6 packs of erasers
I hope this helps. Let me know if you have any questions!!
Cause if its only 19 hours worked is that the whole week? in that case itd be 19*5.75 = 109.25
After running for 18 minutes, Julissa completes 2 kilometers. If she is running a 10-kilometer race at a constant pace, then she comletes 1 kilometer after running 9 minutes. Now you can find the distance she run after 1 minute. This is 
km.
Let t be the time in minutes. Then k, the number of kilometers, can be found from proportion:
- 1 minute
k km - t minutes.
Thus,
Answer: 
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
