Answer : ) 1. which number is not the sum of three consecutive integers? a. 65 b. 150 c. 42 d. 572. train a leaves a station traveling at 48 kilometers per hour. four hours later, train b leaves the same station traveling in the same direction at 68 kilometers per hour. what equation can you use to find the time t in hours that it took train b to catch up with train a, and what is the value of t? a. 68(t – 4) = 48t; t = 13.6 hoursb. 48(t + 4) = 68t; t = 10.6 hoursc. 68(t + 4) = 48t; t = –9.6 hoursd. 48(t + 4) = 68t; t = 9.6 hours3. the sum of two consecutive integers is –27. if n represents one of the integers, what equation can you write to find the value of the two integers and what are the integers? a. n + (n – 1) = –27; –14 and –15b. n + (n + 1) = –27; –14 and –12c. n + (n + 1) = –27; –14 and –13d. n + (n + 1) = –27; –14 and –154. the length of a rectangle is one fourth its width. the perimeter of the rectangle is 60 centimeters. if w represents the width of this rectangle, which expression represents its length? a. l = 1 over 4wb. l = 4wc. l = 60 – wd. l = 60 – 2w

1. A

2. A

3. C

4. A

Step-by-step explanation:

1. Three consecutive integers are represented as n, n+1, and n+2. The sum

n+(n+1)+(n+2) = # must be only true using integers (no fractions/decimals).

So try the first answer 65.

n+(n+1)+(n+2) =65

3n+3=65

3n=62

n=62/3

n=20.6666666666666

This is not an integer so A is the solution.

2. The first train arrives at 48t since it travels 48 km per hour for t hours.

The second train is 68(t-4) since it travels 68 km per hour for the same t hours minus 4 (it left 4 hours later). The time they reach each other is when they are equal. So 68(t-4)=48t or answer A.

3. Two consecutive integers are represented as n and n+1. Notice when added they must equal -27. The only answer option that works is -14 + -13 = -27. Answer C is the solution.

4. The length is one fourth of its width is multiplication by 1/4. So l = 1/4 w or answer A is the solution.

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

i will u if u answer my question in math