There are 16 ten dollar bills and 48 one dollar bills: ($160+$48=$208) and (16+48=64).
Answer:
39
Step-by-step explanation:
not right just need points
The solutions to the problem are 2 and -5.
<h2>Given to us</h2>
![12^{x^2+5x-4}= 12^{2x+6}](https://tex.z-dn.net/?f=12%5E%7Bx%5E2%2B5x-4%7D%3D%2012%5E%7B2x%2B6%7D)
<h2>To find</h2>
the value of x.
<h3>Solution</h3>
![12^{x^2+5x-4}= 12^{2x+6}](https://tex.z-dn.net/?f=12%5E%7Bx%5E2%2B5x-4%7D%3D%2012%5E%7B2x%2B6%7D)
as the base on both, the sides are the same, therefore, equating the powers to each other.
![{x^2+5x-4}= {2x+6}\\x^2 + 5x-2x-4-6=0\\x^2 + 5x-2x-10=0\\x(x+5)-2(x+5)=0\\(x-2)(x+5) = 0\\](https://tex.z-dn.net/?f=%7Bx%5E2%2B5x-4%7D%3D%20%7B2x%2B6%7D%5C%5Cx%5E2%20%2B%205x-2x-4-6%3D0%5C%5Cx%5E2%20%2B%205x-2x-10%3D0%5C%5Cx%28x%2B5%29-2%28x%2B5%29%3D0%5C%5C%28x-2%29%28x%2B5%29%20%3D%200%5C%5C)
we got the two factors.
Now equating both the factors against 0,
![(x-2) =0\\x=2](https://tex.z-dn.net/?f=%28x-2%29%20%3D0%5C%5Cx%3D2)
![(x+5)=0\\x=-5](https://tex.z-dn.net/?f=%28x%2B5%29%3D0%5C%5Cx%3D-5)
Therefore, the solutions to the problem are 2 and -5.
Learn more about factorization:
brainly.com/question/20293447?referrer=searchResults
7(3)^2 - 3(3) - 6 = 7*3*3 - 3*3 - 6 = 63 - 9 - 6 = 48
Step-by-step explanation:
kindly find attached a table of how revenue relates to price.
i is used for count increments of $15.
P=200-15i
COST for flying 1 person was given at $100, for ONE PERSON.
KEY IDEA: P=r-c, PROFIT is revenue minus cost, but understand the cost will be for all ticket buyers who are to fly.
As your description gives, P is for-profit and p is for the price of a ticket.
(a) profit in terms of number of tickets sold;
p=r-c
n for a count of tickets
p=r-c*n
p=(200-15i)n-100n
No way getting away from the unknown ticket price. Otherwise, you could simply say
p=pn-100n