Simple algebra equation .

an adult ticket to the amusement park costs $24.95 and a child’s ticket costs $15.95. a group of 10 people paid $186.50 to enter the park. how many were adults?
this is my equation:

24.95x+ 15.95y
10 = 186.50

i don't know what to do from here.

There are 3 adult tickets and 7 child tickets

Step-by-step explanation:

You are very close.

Let x = the number of adult tickets

Let y = number of child tickets

24.95x+ 15.95y = 186.50

The total number of tickets is 10

x+y =10

Subtract y from each side

x+y-y = 10-y

x =10-y

Substitute this into the first equation

24.95x+ 15.95y = 186.50

24.95(10-y) +15.95y = 186.50

Distribute

249.5 - 24.95y +15.95y =186.5

Combine like terms

249.5 - 9y = 186.50

Subtract 249.5 from each side

249.5-249.5 - 9y = 186.50-249.5

-9y =-63

Divide each side by -9

-9y/-9 = -63/-9

y =7

Now we need to find x

x+y =10

x+7 =10

Subtract 7 from each side

x+7-7 =10-7

x =3

There are 3 adult tickets and 7 child tickets

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}

3; 9 is 8 to -4

step-by-step explanation: