9514 1404 393
Answer:
5 hours
Step-by-step explanation:
A quick way to look at this is to compare the difference in hourly charge to the difference in 0-hour charge.
The first day, the charge is $3 more than $12 per hour.
The second day, the charge is $12 less than $15 per hour.
The difference in 0-hour charges is 3 -(-12) = 15. The difference in per-hour charges is 15 -12 = 3. The ratio of these is ...
$15/($3/h) = 5 h
The charges are the same after 5 hours.
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If you write equations for the charges, they will look like ...
y1 = 15 + 12(x -1)
y2 = 3 + 15(x -1)
Equating these charges, we have ...
15 +12(x -1) = 3 + 15(x -1)
12x +3 = 15x -12 . . . . . . . . eliminate parentheses
15 = 3x . . . . . . . . . . add 12-12x
x = 15/3 = 5 . . . . . . divide by 3
You might notice that the math here is very similar to that described in words, above.
The charges are the same after 5 hours.
Answer:
There are 45 2-point questions and 5 4-point questions
Step-by-step explanation:
Answer:
The larger acute angle is equal to 50.8 degrees.
Step-by-step explanation:
Let's solve for both of the acute angles for the purpose of checking our work at the end with angle A being the top angle and angle B being the one on the base of the triangle (that's not the 90 degrees one). Determining whether to use sin/cos/tan comes from SOH-CAH-TOA.
A = cos^-1 (2√6/2√15)
However, you need to move the radical out of the denominator by multiplying √15 to the numerator and denominator. You should come up with (2√90)/30. So,
A = cos^-1 (2√90/30) = 50.768 degrees.
B = sin^-1 (2√90/30) = 39.231 degrees.
Now, we can check the work by adding the 2 angles to 90 and, if it comes to 180, it's right.
cos^-1 (2√90/30) + sin^-1 (2√90/30) + 90 = 180.
If you have any questions on where I got a formula or any step, feel free to ask in the comments!