We can set up 2 equations given the information.
Let the price of advance ticket be represented by A and same day ticket by S
A + S = 50
20A + 40S = 1700
Solve for A in the first equation by subtracting S on both sides.
U will get A = 50 - S
Now substitute 50 - S for A in the second equation.
20 (50 - S) + 40S = 1700
1000 - 20S + 40S = 1700
20S = 700
S = 35
Same day ticket costs $35 and advance ticket costs $15
The answer is A. To understand that, you need to put it in order by the x's and the y's to get x^2-6x + y^2-16y = -48. Now complete the square on both the x and the y terms to get (x^2-6x+9) +(y^2-16y+64) = -48+9+64. Rewriting that in vertex form on the left and doing the math on the right gives you
(x-3)^2 + (y-8)^2 = 25, which shows you a center of (3,8) and a radius of 5.
Answer:
Domain is -3
Step-by-step explanation:
Given that coordinate of a graph is (-3, -2)
We have to find out the domain.
We know that if A and B are two sets, a mapping from A to B is the subset of cartesian product AxB.
Domain is the set of values of A which have images in B.
Use the above definition.
We have A = {-3,...} and B = {-2,....}
The mapping is from -3 to -2
Hence domain is -3
Answer:
The number of distinct arrangements is <em>12600</em><em>.</em>
Step-by-step explanation:
This is a permutation type of question and therefore the number of distinguishable permutations is:
n!/(n₁! n₂! n₃! ... nₓ!)
where
- n₁, n₂, n₃ ... is the number of arrangements for each object
- n is the number of objects
- nₓ is the number of arrangements for the last object
In this case
- n₁ is the identical copies of Hamlet
- n₂ is the identical copies of Macbeth
- n₃ is the identical copies of Romeo and Juliet
- nₓ = n₄ is the one copy of Midsummer's Night Dream
Therefore,
<em>Number of distinct arrangements = 10!/(4! × 3! × 2! × 1!)</em>
<em> = </em><em>12600 ways</em>
<em />
Thus, the number of distinct arrangements is <em>12600</em><em>.</em>
Answer: No she'll need atlease 9 dollars
Step-by-step explanation:
48.95 x 18% is 8.811 which is greater than
