6w-10 represents the total amount of fence will need to complete the rectangle of fence in his yard. The fence will need two lengths and 2 widths to complete the rectangle. If the width is w, and the length is 2w-5, then the total for the rectangle would be w+w+2w-5+2w-5 which simplifies to 6w-10.
Answer:
Theres no picture?
Step-by-step explanation:
Answer:
The correct evaluation is 1 1/3 (one and one-third) and not 16 and one-third
Step-by-step explanation:
The student was wrong in his evaluation because the correct result should be 1 1/3 (one and one-third) and not 16 and one-third
The expression '1/3 more than the product of four and a number' means
(4g + 1/3)
Evaluating the expression when g = 1/4
You will have
4×1/4 + 1/3
= 1/1 + 1/3
Find the LCM of 1 and 3 and add
= (3+1)/3
=4/3
= 1 1/3
The correct evaluation is 1 1/3 (one and one-third) and not 16 and one-third
Answer:
y = 15x+32
Step-by-step explanation:
Given
y to be the total minutes Andrew will play video games today
x is the total number of games played after school
If he wants to play an additional 15 minutes each of x games after school today, the total minutes he will use for the x games will be 15* x = (15x)minutes
We are told that he has initially spent 32minutes on video games. hence;
Total minutes Andrew used on games y = (15*x)+32
The required equation will be y = 15x+32
<u>Answer:</u>
Cost of package of paper = 4$
Cost of stapler = 7$
<u>Explanation:</u>
Consider the cost of package of paper = x and that of stapler = y.
Now, we are given that cost of 3 paper packages and 4 staplers = 40$
Hence we get, 3x + 4y = 40 as 1st equation.
we are also given, cost of 5 paper packages and 6 staplers = 62$
Hence, the second equation is 5x + 6y = 62
Now, solving the two equations by method of elimination, we first equate coefficients of any one variable say x by multiplying 1st equation by 5 and second by 3 we get ->
15x + 20y = 200
15x + 18y= 186
Subtracting the two we get y = 7 and substituting this value of y in first equation we get x = 4
which gives the required cost of one paper package = x = 4$
and one stapler = y = 7$
