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Oksana_A [137]
3 years ago
10

14. The table shows the survey results of exercises

Mathematics
1 answer:
Akimi4 [234]3 years ago
3 0

Answer:

Answers in the question are A, B, and E.

The answers to the table are:

         yes     no     total

yes:     12      25      37

no:      12       6        18

total:   24      31       55

Hopefully this helped, sorry if I was too late :)

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Where would fraction<br> 22/11 belong on a #6 number line
pickupchik [31]
22/11 is equal to 2
So, it would be where 2 is on the number line.
6 0
3 years ago
WILL GIVE BRAINLIEST!!!!
boyakko [2]

Exponential functions are known to increase geometrically. An example of exponential function is  p(x) = 500(1.02)^x

<h3>Exponential functions</h3>

Exponential functions are known to increase geometrically. The standard exponential function is given as:

y = ab^x

a is the base

x is the exponent

From the given options, the function written in this form is

p(x) = 500(1.02)^x. Hence an example of exponential function is

p(x) = 500(1.02)^x

Learn more on exponential function here: brainly.com/question/12940982

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8 0
2 years ago
(4 ten Thousands 4 thousands) x 10 =
sweet-ann [11.9K]
The answer would be

40,000
8 0
3 years ago
Read 2 more answers
Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
1 year ago
What is the answer to -5(6+9k)
anzhelika [568]

Answer: The answer is -30 - 45 k

Step-by-step explanation:

8 0
3 years ago
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