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Fiesta28 [93]
3 years ago
14

How do I solve 4 2/5 - 4 3/4

Mathematics
1 answer:
snow_lady [41]3 years ago
5 0
Put the denominators on 20 and then it will be easier for you to soundtrack them
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Enter the y coordinate of the solution to this system of equations -2x+3y=-6. 5x-6y=15
adoni [48]
The intersect of the 2 lines is (3,0) and the y coordinate is 0
4 0
4 years ago
A group of 328 people consists of men ,women, and children. There are five times as many men than children and twice as many wom
maxonik [38]

Step-by-step explanation:

Ultimately we know:

Men = * 5

Women = * 2

Children =

We can already make the estimation that the amount of children will be below 50. So by trial and error you will get to your answer by substituting with 41.

Children - 41

Women - 41 x 2 = 82

Men - 41 x 5 = 205

205 + 41 + 82 = 328

So your answer is <u>41</u>

5 0
2 years ago
Bob traveled 90 miles which was 83 1/3 % of his trip. How long was the trip?
lilavasa [31]
Let x be total amount traveled Therefore, x(.8333) = 90 x = 108 108 miles

hope i helped
5 0
3 years ago
Read 2 more answers
A toolbox has 10 screwdrivers Sid 6 wrenches.
SVEN [57.7K]

Answer: There are 4 more wrenches in the toolbox then the screwdrivers.

Step-by-step explanation: Add the 6 wrenches Sid put in the toolbox with the 8 wrenches Bella added to get 14 wrenches in total. Then, subtract the 10 screwdrivers from the 14 wrenches to get 4 wrenches.

8 0
4 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
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