So that equation was definitely correct...
When you expand the equation in the bracket you'll find out that you'll get a^6 + 4a^4 + !6a^2 - 4a^4 - 16a^2 -64. then your final result will be a^6 - 64
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Answer:
its 4
Step-by-step explanation:
1 2 3 4
hope i helped
Answer:
A or the top option
Step-by-step explanation:
I used process of elimination
B simply does not work because in order for 4 to be subtracted by another number to equal 18, that number (width) would have to be negative
C also isn't correct since if it was solved out, we would add four to both sides, making the left side equal to w/2 + 4 which isn't what we want because the length is 4 shorter, not longer
D can be dealt with in the same way, by adding w/2 on both sides, making the left side 4 + w/2, the same as the previous option.
Answer:
- 280 student tickets
- 520 adult tickets
Step-by-step explanation:
You may recognize that you are given two relationships between two unknowns. You can write equations for that.
You are asked for numbers of adult tickets and of student tickets. It often works well to let the values you're asked for be represented by variables. We can choose "a" for the number of adult tickets, and "s" for the number of student tickets. Then the problem statement tells us the relationships ...
a + s = 800 . . . . . . 800 tickets were sold
12.50a + 7.50s = 8600 . . . . . . . revenue from sales was 8600
(You are supposed to know that the revenue from selling "a" adult tickets is found by multiplying the ticket price by the number of tickets: 12.50a.)
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You can solve these two equations any number of ways. One way is to do it by <em>elimination</em>. We can multiply the first equation by 12.50 and subtract the second equation:
12.50(a +s) -(12.50a +7.50s) = 12.50(800) -(8600)
5s = 1400 . . . . simplify. (The "a" variable has been eliminated.)
s = 280 . . . . . . divide by 5
Then the number of adult tickets can be found from the first equation:
a + 280 = 800
a = 520
280 student tickets and 520 adult tickets were sold.
