Mr. Kaplan bought 11 tickets to the circus and spent $50. He bought child tickets for $4 each and bought adult tickets for $7 each. How many child and adult tickets did Mr. Kaplan buy?

Mr Kaplan bought 9 child tickets and 2 adult tickets.

Step-by-step explanation:

- Mr Kaplan bought 11 tickets

- He spent $50

- Child tickets cost $4 each

- Adult tickets cost $4 each.

We want to know how many adult ticket and child ticket Mr Kaplan bought.

Better still, we want to know exactly how many $4's and $7's make $50.

Let us try out the numbers that sum up to 11.

10 + 1 = 11

9 + 2 = 11

8 + 3 = 11

7 + 4 = 11

6 + 5 = 11

Next, let us check which one(s) of these number when applied to the costs of tickets gives 50

(10×$4) + (1 × $7) = $47 ≠ $50

(10×$7) + (1×$4) = $74 ≠ 50 (10×$7, 9×$7, 8×$7, and 7×$7 are already too much, it is reasonable to ignore them)

Let's try the next one

(9×$4) + (2×$7) = $50 (This works)

Next,

(8×$4) + (3×$7) = $53 ≠ 50

(7×$4) + (4×$7) = $56 ≠ 50

(6×$5) + (5×$7) = $65 ≠ 50

So, we can conclude that Mr Kaplan bought 9 child tickets and 2 adult tickets.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

98%

step-by-step explanation: