W equals 7 :) you’re welcome
Your answer is A. eastings
Let me know if you have another question ☺
We know that $375 is spent for three weeks period
One week period is 5 days, so for three weeks periods, there are 15 days in total when the truck use the gasoline.
The amount spent on gasoline for one day is $375÷15 = $25
Back in my old days, everyone HATED fractions!! So in order to rid of them, you have to multiply them with the greatest common factor in all of them (GCF). So it would look like this:
GCF of 3,6, and 6 would be 6.
Once you know that, you multiply it to the whole equation.
6(1/3z +1/6 = 5/6)
Then do the math and you should be left over with no fractions at all.
2z+1=5
If you don't know your algebra 1 skills, you must move the numbers to one side and keep the variables on the other side -> 2z=4
•Divide by 2 on each side
z=2 will be the answer if you want to practice :)
The answer to the question ->
{multiply by 6}.
If it is decimals, you MUST want to take away that annoying dot right there just by multiplying by 10,100,1000,10,000 and so on for the whole equation.
Pretend we have 1.307
We multiply by 1000 because 7 is in the thousandths place so you get a whole number.
-> 1307!
So the problem asks
3.7 + 2.75k = 27.35
You see that the GCF can be only by 10, 100 and so on because you don't want the dots. Then the GCF in this case would be 100.
So {multiply by 100}.
100(3.7 + 2.75k = 27.35)
Changed to 370+275k=2735
•Subtract the numbers
275k=2355
•Divide
k=8.6
Hope this helps!
Answer:
The correct answer is B.
Step-by-step explanation:
We have the expression:
In this case, B is the correct answer.
B states that any sample point can be chosen without affecting the limit.
This is a true statement. Regardless of which point we choose for our sample point, since n approaches infinity, we will have infinite partitions.
Hence, this will not affect the sum of our expression.
A and C are false because we have infinite partitions, so the statements cannot be true. D is not true simply because the left-hand values do not have to be the same as the midpoint-values depending on the behavior of the function for [a, b].
