The school sold 200 tickets to play. Student tickets are $4 and adult tickets are $6. If they made a total $930 how many of each
1 answer:
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Answer:
See attached photo for graph
Step-by-step explanation:
Solve the equation for y... (that means get y by itself)
4y = -5x - 6 (subtract 5x from both sides)
y = -(5/4)x - 6/4 (divide both sides by 4)
y = -(5/4)x - 3/2 (simplify)
Now you can graph using slope-intercept method. The y-intercept is -3/2, and the slope is -5/4
Answer:
1980
We can choose 2 men from 9 in 36 in ways.
We can choose 2 women from 11 in 55 ways.
For any arrangement of 2 men there are 55 ways of combining them with 2 women => 55 * 36 = 1980 total ways.
If you're wondering how i got the numbers i used the combination formula:
C(n, k) =
Answer: fourth option
Explanation:
1) the pair x = 3 f(x) = 0, leads you to probe this:
f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0
=> A [4^2] = - C
A[16] = - C
if A = 1/4
16 / 4 = 4 => C = - 4
That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4
2) Now you verify the images for that function for all the x-values of the table:
x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check
x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check
x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.
Therefore, you have proved that the answer is the fourth option.
Explanation:
1) the pair x = 3 f(x) = 0, leads you to probe this:
f(3) = 0 = A [4 ^ (3 - 1) ] + C = 0
=> A [4^2] = - C
A[16] = - C
if A = 1/4
16 / 4 = 4 => C = - 4
That leads you to the function f(x) = [1/4] 4 ^( x - 1) - 4
2) Now you verify the images for that function for all the x-values of the table:
x = 2 => f(2) + [1/4] 4 ^ (2 - 1) - 4 = [1/4] 4 - 4 = 4 / 4 - 4 = 1 - 4 = - 3 => check
x = 3 => f(3) = [1/4] 4^ (3 - 1) - 4 = [1/4] 4^2 - 4 = 16 / 4 - 4 = 4 - 4 = 0 => check
x = 4 -> f(4) = [1/4] 4^ (4-1) - 4 = [1/4] 4^(3) - 4 = (4^3) / 4 - 4 = 4^2 - 4 = 16 - 4 = 12 => check.
Therefore, you have proved that the answer is the fourth option.