Shayla tutors middle school students and high school students. She charges a different rate for each. Last week, she tutored middle school students for 5 hours and high school students for 12 hours. This week, she tutored middle school students for 6 hours and high school students for 10 hours. She earned $400 last week and $370 this week. How much does she charge each?

Shayla charges $20 per hour for middle school students and $25 per hour for high school students.

Step-by-step explanation:

System of Equations

Let's call:

x = Shayla's hourly charge for middle school students

y = Shayla's hourly charge for high school students

Last week, Shayla tutored middle school students for 5 hours and high school students for 12 hours. The total earning was 5x+12y. She earned $400 last week, thus:

5x + 12y = 400        [1]

This week, she tutored middle school students for 6 hours and high school students for 10 hours. The total earnings were 6x+10y and it represented $370, thus:

6x + 10y = 370

Dividing by 2:

3x + 5y = 185        [2]

We have formed the system of equations:

5x + 12y = 400        [1]

3x + 5y = 185        [2]

Multiply [1] by -3 and [2] by 5:

-15x - 36y = -1200

15x + 25y = 925

Adding both equations:

-11y = -275

Dividing by -11:

y = -275/(-11) = 25

y = 25

Substituting in [1]:

5x + 12*25 = 400

5x + 300 = 400

5x = 400 - 300 = 100

x = 100/5 = 20

x = 20

Shayla charges $20 per hour for middle school students and $25 per hour for high school students.

step-by-step explanation:

i think you should think about the question and not look online.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}