I need help with these 3 problems.

1. An amusement park charges admission plus a fee for each ride you go on. 2 admissions and 5 rides cost $22. 2 admissions and X rides cost $24. What is the charge for 1 admission and the charge for 1 ride.

2. Find the lenght of the hypotenuse of a right triangle with legs of length 6 and 10. Give you answer in simplest radical form.

3. Solve. 3x^2-5x-1=2

PLEASE HELP ME!!!

1. Where it says X rides, are you sure there is not a number there? There is not enough information in that problem the way it is written.

2. For right triangles, use the pythagorean theorem. a^2+b^2=c^2..6^2+10^2=? Then simplify that.
3. Use the quadratic formula: (-b+-sqrtb^2-4ac)/2a
Post your answers if you'd like them checked.

2a+5r=22

2a + xr = 24
I need to know x to solve for a and r.

h^2 = 6^2 + 10^2
h^2 = 136
h = 2 sqrt 34

x = [5 +/- sqrt(25+12)]/6
= [ 5 +/- sqrt (37)]/6

on questin 1 the number was cut off the page i guessed it was either 6 or 7

here are the answers i got please check them.

1. Still need help
2. 2 sqrt14
3.5+-sqrt1/6

at a travel agency the cost of a trip to mexico is $350 for an adult and $200 for a child. One month, the agency sold 50 trips. Of these trips,y trips were for children. The travel agency collected c dollars from selling all 50 trips. Write A linear equation for C in terms of y.

Of course, I'd be happy to help you with these problems! Let's take them one at a time.

Problem 1:
To solve this problem, let's assign some variables. Let's say the cost for one admission is A dollars and the cost for one ride is R dollars.

According to the given information: 2 admissions + 5 rides cost $22.
So we can write the equation: 2A + 5R = 22.

Similarly, we're told that 2 admissions + X rides cost $24.
So the second equation is: 2A + XR = 24.

To find the charges for 1 admission and 1 ride, we need to solve these two equations simultaneously.

One way to do this is by using the method of elimination. Let's eliminate variable A by multiplying the first equation by 2 and the second equation by -2:

(2 * 2A) + (2 * 5R) = 2 * 22
(-2 * 2A) + (-2 * XR) = -2 * 24

This simplifies to:
4A + 10R = 44
-4A - 2XR = -48

Adding both equations together, we eliminate variable A:
(4A + 10R) + (-4A - 2XR) = 44 + (-48)
8R - 2XR = -4

Now, let's isolate R on one side of the equation by factoring out the common term R:
R(8 - 2X) = -4

Divide both sides of the equation by (8 - 2X):
R = -4 / (8 - 2X)

So, the charge for 1 ride is -4 divided by the quantity (8 minus 2 times X).

To find the charge for 1 admission, substitute the value of R from the first equation (2A + 5R = 22) into the equation we found for R. Solving for A, we get:
2A = 22 - 5R
A = (22 - 5R) / 2

Therefore, the charge for 1 admission is (22 minus 5 times R) divided by 2.

Problem 2:
To find the length of the hypotenuse of a right triangle, we can use the Pythagorean theorem. The theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

In this case, the two legs of the triangle have lengths of 6 and 10.
So, by applying Pythagoras' theorem, we have:

Hypotenuse^2 = 6^2 + 10^2
Hypotenuse^2 = 36 + 100
Hypotenuse^2 = 136

To simplify the radical form of the solution, we look for the perfect square factors of 136.
The prime factorization of 136 is 2^3 * 17.

Therefore, the square root of 136 can be simplified as:
√136 = √(2^2 * 2 * 17)
= 2√(2 * 17)
= 2√34

Therefore, the length of the hypotenuse of the right triangle is represented as 2√34.

Problem 3:
To solve the equation 3x^2 - 5x - 1 = 2, we'll first move all the terms to one side to set it equal to zero:

3x^2 - 5x - 1 - 2 = 0
3x^2 - 5x - 3 = 0

This is a quadratic equation, so we can solve it using various methods such as factoring, completing the square, or using the quadratic formula.

In this case, let's use the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solution for x can be found using the formula:

x = [-b ± √(b^2 - 4ac)] / (2a)

In our equation, a = 3, b = -5, and c = -3.

Plugging these values into the quadratic formula, we get:

x = [-(5) ± √((-5)^2 - 4(3)(-3))] / (2(3))
x = [-(-5) ± √(25 + 36)] / 6
x = [5 ± √61] / 6

So the solutions to the equation are (5 + √61) / 6 and (5 - √61) / 6.

I hope this helps you with your problems! If you have any further questions, feel free to ask.