Answer: See below
Step-by-step explanation:
27. -(a-3)
28. (b-1)(b+3)
29. (c+4)(c+5)
30. d(d+5)
31. -(3/4)(2e-5)
Sorry - I don't have time to enter the details. Look for areas where the expressions can be factored in a manner that forms as many equivalent expressions in both the numerator and denominator.
For example: In problem 30:
(5d-20)/(d^2+d-20) * [??]/20d = 1/4
Factor:
<u>(5(d-4))</u> <u>d(d+5)</u> = 1/4
(d-4)(d+5<u>)</u> 20d
The (d-4), d+5, and d terms cancel, leaving
5/20 = 1/4
Answer:
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Answer:
A. 50
Step-by-step explanation:
Answer:
The north campus had 
while the south campus had 
.
Step-by-step explanation:
The question stated that there are 
students in the merged campus. That's the sum of the number of students in the north and the south campus before the merger.
Let 
denote the number of students in the north campus before the merger. The other 
students would all belong to the south campus before the merger.
Before the merger, 
of the students of the north campus are of music majors. In terms of , that's 
students.
On the other hand, 
of the students of the south campus are of music majors. That corresponds to 
students.
With a similar logic, the number of music students in the merged east campus will be 
.
The question implies that the sum of students in the two campuses should be equal to the number of music students in the east campus. That is:
.
Solve for :
.
In other words, there are students in the north campus and 
students in the south campus before the merger.
Answer:
Domain: t≥0
Range:h(t) ≤ 0.85
Step-by-step explanation:
The domain of a function is all the valid values you can enter into the function. Because time can't be negative the domain is t>0. The range or h(t) is the result, because the object can't go through the ground, it can't be less than zero. 0≤h(t)
0≤-4.87t^2+18.75t
0≤-4.87t+18.75
-18.75≤-4.87t
which is approx.: 0.85≥t or t≤0.85
