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Pavlova-9 [17]
3 years ago
13

PLEASE HELP THIS IS VERY IMPORTANT I WILL GIVE U BRAINLIEST AND 5 STAR AND THANKS

Mathematics
1 answer:
slega [8]3 years ago
6 0

Answer:

it  wont let me put any numbers down, it keeps saying they're improper

Step-by-step explanation:

You might be interested in
What is the value of 500$ invested at 4% interest compounded annually for 7 years
DiKsa [7]

Answer:

657.96

Step-by-step explanation:

use formula A=P(1+r/n)^nt

A=500(1+.04/1)^1*7

A=500(1.04)^7

A=500(1.3159~)

A= 657.96~

5 0
3 years ago
Trevor pays $125 in fees each semester plus $650 per college course.
slava [35]
I hope this helps you



for per college course $650


650.?+125=2725
7 0
3 years ago
what is the dividend yield on stock A that sells at $15/share when company A pays a quarterly dividend of $0.05 per share
polet [3.4K]

Answer:

300

Step-by-step explanation:

15/0.05=300

3 0
3 years ago
Please help! I've already answered part a, I don't understand what part b is asking.
Luba_88 [7]

Step-by-step explanation:

So, there is something known as a removable discontinuity, and it's essentially where you can define f(x) using the most simplified fraction, where you could normally not define f(x).

So we have the following equation:

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

As you may know, we cannot divide a number by the value of zero. When the denominator is equal to zero, on the graph this will appear as a vertical asymptote, where x approaches the value that makes the denominator zero, but never actually reaches it.

If you look at each denominator, you can set them equal to zero to find the vertical asymptotes

x+1 = 0

x=-1

There should be a vertical asymptote at x=-1, since it would make two of the denominators equal to -1, but let's divide the two fractions first.

Original Equation

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

Keep, change, flip

f(x) = (\frac{x+5}{x+1}*\frac{(x-4)(x+1)}{(x+3)(x-2)})-\frac{1}{x-2}

Multiply the two fractions

f(x) = (\frac{(x+5)(x-4)(x+1)}{(x+1)(x+3)(x-2)})-\frac{1}{x-2}

Notice how the x+1 is in the numerator and fraction? That means we can cancel it out!

f(x) = (\frac{(x+5)(x-4)}{(x+3)(x-2)})-\frac{1}{x-2}

In this simplified version of the fraction, we can technically define f(-1), but in the original version, since it's not defined there is a removable discontinuity at x=-1, meaning there is no vertical asymptote, but the function is still not defined at f(-1), and there will be a hole at that point.

4 0
2 years ago
In a quiz its just middle school work so PLS help me
Tems11 [23]

Answer:

5.25x - 6

Step-by-step explanation:

<u>1.75x</u> -2 + <u>1.75x</u> -2 + <u>1.75x </u>-2

5.25x <u>- 2 - 2 - 2</u>

5.25x - 6

8 0
3 years ago
Read 2 more answers
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