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atroni [7]
4 years ago
8

What is 25/30 simplified(25/30 is a fraction)

Mathematics
2 answers:
Irina-Kira [14]4 years ago
8 0
It's 4/3 poiojehnemejsiiekeje
dybincka [34]4 years ago
7 0
You first have to find a multiple that can divide both numbers equally. 25 divided by 5 is 5. And 30 divided by 5 equals 6. So the simplified fraction of 25/30 is 5/6
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3 years ago
Simplify: square root 108+square root 125-square root 75​
navik [9.2K]

Answer:

11\sqrt{3} + 5\sqrt{5}

Step-by-step explanation:

\sqrt{108}+\sqrt{125} - \sqrt75}

1. \sqrt{108}  = 6\sqrt{3}\

2. \sqrt{125} = 5\sqrt{5}

3. \sqrt{75} =  5\sqrt{3}

4.    Add 6\sqrt{3} + 5\sqrt{3}  = 11\sqrt{3}

your left with 11\sqrt{3}+5\sqrt{5}

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4 years ago
The sum of 48 and twice a number is 58. find the number
GalinKa [24]

Answer:

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Step-by-step explanation:

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2 years ago
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Write out each step of the 3-step test for continuity for the following functions at the given point. If the function is discont
balandron [24]

If we simply substitute the value x=2 in the expression we have

f(2) = \dfrac{4-12+8}{4-2-2} = \dfrac{0}{0}

which is undefined.

But if we factor both numerator and denominator, we have

f(x)=\dfrac{x^2-6x+8}{x^2-x-2} = \dfrac{(x-2)(x-4)}{(x+1)(x-2)}

Since we are studying the limit as x approaches 2, we can assume that x is not 2. In this case, we can simplify the (x-2) parenthesis, and the expression becomes

f(x)=\dfrac{x-4}{x+1}

And we can evaluate this at 2 with no problems:

f(2) = \dfrac{2-4}{2+1} = -\dfrac{2}{3}

So, we have

\displaystyle \lim_{x\to 2}f(x) = -\dfrac{2}{3}

This means that in this case both left and right limits exist and are the same, so the limit exists, but the function is not defined at x=2. This is a removable discontinuity, because we can define the function as its limit, and we have a continuous function at x=2:

f(x) = \begin{cases}\dfrac{x^2-6x+8}{x^2-x-2} &\text{if }x \neq 2\\ -\frac{2}{3} &\text{if }x = 2\end{cases}

8 0
4 years ago
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