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Sliva [168]
3 years ago
14

many colleges determine the number of full time students f,by using the formula f=n/15 where n is the total number of credits fo

r each semester if a student registers for 46140 credits how many full time students does the campus have. then what is the formula for n  n=
Mathematics
1 answer:
Cloud [144]3 years ago
7 0
"A student" doesn't register for 46,140 credits, unless he wants to be very very busy.

If ALL of the students on campus register for 46,140 credits, then many colleges
would estimate the number of full-time students by using the formula

f = n / 15 = 46,140 / 15 = <u>3,076</u> students.

That's not the formula for 'n'.  'n' is the total number of credits for the semester,
and the question tells you what 'n' is.  Knowing 'n', the formula helps you find 'f'.
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One more than the difference between 18 and seven times in number is -9
almond37 [142]
18-7n+1= -9

You just have to ignore the first thing it asks for, and then write out what it's saying. Since it's asking for the difference, you have to subtract what it tells you to. In this case they are asking for you to subtract 7 times a number from 18 which is the same as writing 18-7n. Then, go back to the first thing it asks for and that wants you to add one. You now have this: 18-7n+1. Now all you have to do is set it equal to whatever it asks you for, which is -9. The final product should look like this:

18-7n+1= -9

I hope this helps and if you could let me know if I helped, that would be greatly appreciated!
7 0
3 years ago
Find the slope of the line through the pair of points. A(2, 3), P(2, 9)
AVprozaik [17]
A(x_1;\ y_1);\ B(x_2;\ y_2)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\-------------------\\\\A(2;\ 3)\to x_1=2\ and\ y_1=3\\B(2;9)\to x_2=2\ and\ y_2=9\\\\m=\dfrac{9-3}{2-2}=\dfrac{7}{0}-divided\ by\ 0\\\\Answer:\ The\ slope\ unde fined. It's\ a\ vertical\ line\ x=2.
8 0
3 years ago
The area of a circle is 81 pi meters squared. What is the circumference, in meters?
pickupchik [31]

Answer:

I think the answer is 42.

6 0
3 years ago
The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f
Nataliya [291]

The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.

Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.

By the fundamental theorem of calculus,

\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so

\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7

\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}

8 0
2 years ago
I need help, along with an explanation, for this question.
Aleksandr-060686 [28]

Answer:

1.  -9

2.  -15

Step-by-step explanation:

Reduce the expression, if possible, by cancelling the common factors.

8 0
3 years ago
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