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grandymaker [24]
3 years ago
10

Turning improper fractions into mixed numbers . Help?

Mathematics
1 answer:
Georgia [21]3 years ago
3 0
A. 5 2/5
b. 32 3/8
c. 2 1/7
d. 6 1/4
 
Hope this helps!
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How to find sample standard deviation from population standard deviation.
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Step-by-step explanation:

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3 years ago
Please tell me what x is 3(x+6)=48
Alekssandra [29.7K]

Answer

X=10

Step by step explanation

48÷3 = 16

16-6 = 10

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4 years ago
Solve for x. x/2≥−4 <br> a. x≥−2 <br> b. x≤−2 <br> c. x≥−8 <br> d. x≤−8
yKpoI14uk [10]
If you are satisfied please like

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What times -4 equals 23?
Lady_Fox [76]
Ask your self what is -2*?=23.
Then times that answer by 2.
5 0
3 years ago
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
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