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stira [4]
3 years ago
15

Express each ratio as a fraction in the simplest form. 12 beetles out of 18 insects .

Mathematics
1 answer:
slava [35]3 years ago
3 0

Answer:

2/3

Step-by-step explanation:

12/6=2

18/6=3

so

the fraction is 2/3

i think this is what your teachers means, lmk if thats wrong

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Answer:

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Step-by-step explanation:

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A basketball player averaged 20 points a game over the course of six games. His
Nataly_w [17]
<h3>Hello There!!</h3>

<h3><u>Given,</u></h3>

Average Points= 20

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3 0
2 years ago
A group of middle school students were randomly chosen and asked about whether or not they will attend the school dance. Of thes
Stels [109]

Answer:

80 Students were surveyed.

Step-by-step explanation:

To find the answer we need to do a Rule of Three. The Rule of Three is a Mathematical Rule that will help us to solve problems based on proportions.

1. I know that a 65% of students are 52 students. I need to know how many students represent that 100%.

2. So if 52 students mean _____ 65%

             X students mean  _____ 100%

Pay attention here. We have to put the same elements in the same columns. Students with studients and percentage with percentage.

3. Now I have to cross these details. I have to multiple 52*100. So:

52*100= 5200

4. Then divide that result by 65.

5200/65= 80

5. So 80 is the total number of students surveyed.

3 0
3 years ago
Let <img src="https://tex.z-dn.net/?f=r%28x%29%20%3D%20%5Cfrac%7B8x-x%5E%7B2%7D%20%7D%7Bx%5E%7B4%7D-64x%5E%7B2%7D%7D" id="TexFor
Reil [10]

Answer:

<em>x=8</em>

Step-by-step explanation:

<u>Discontinuity of a Function</u>

We can find some functions whose graphs cannot be plotted in one stroke. It can be a hole or a vertical asymptote or a jump. To find a possible hole in a rational function, we must set both numerator and denominator to 0 independently. If a common point is found, it's a candidate for a hole if the function could eventually be redefined as continuous.

Let's find the zeros of the numerator

8x-x^2=0

Factoring

x(8-x)=0

We find two solutions: x=0, x=8

Let's find the zeros of the denominator

x^4-64x^2=0

Factoring

x^2(x-8)(x+8)=0

We find three roots: x=0, x=8, x=-8

There are two common points where the function can have holes, those are

x=0,\ x=8

We are not sure if those values are holes or not until we find the limits

\displaystyle \lim\limits_{x \rightarrow 8}\frac{x(8-x)}{x^2(x-8)(x+8)}

Simplifying

\displaystyle =\lim\limits_{x \rightarrow 8}-\frac{1}{x(x+8)}

\displaystyle =-\frac{1}{128}

Since the limit exists, the function can be redefined to cover up the hole. Now let's find the limit in x=0

\displaystyle \lim\limits_{x \rightarrow 0}\frac{x(8-x)}{x^2(x-8)(x+8)}

Simplifying

\displaystyle =\lim\limits_{x \rightarrow 0}-\frac{1}{x(x+8)}

\displaystyle =-\frac{1}{0}=-\infty

The limit does not exist and goes to infinity, it's not a hole, thus the only hole occurs when x=8

5 0
3 years ago
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