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2 years ago

Plzz help meeee plzzzz help

Mathematics

1 answer:

ruslelena [56]2 years ago

8 0

Answer: A. 20%

Step-by-step explanation:

First, to find the total number of teachers, we can add 25+32+18+5+20=100

Next, we divide the number of teachers in 2014 (20) by the total number of teachers (100) as follows:

20 ÷ 100 = 0.2

Finally, to make 0.2 a percentage, we can multiply 0.2 X 100 = 20

Therefore, the correct answer is A.

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A side of the triangle below has been extended to form an exterior angle of 132°. Find the value of X

Zinaida [17]

Answer:

x = 114°

Step-by-step explanation:

The exterior angle of a triangle is equal to the sum of the 2 opposite interior angles, thus

x + 18° = 132° ( subtract 18° from both sides )

x = 114°

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2 years ago

Suppose f(x) = 6x^2 + 2x − 7 and g(x) = 4x − 3. Find each of the following functions.

FinnZ [79.3K]

Step-by-step explanation:

(f+g)(x) ➡ f(x) + g(x) ➡ 6x^2 + 2x -7 + 4x - 3 =6x^2 + 6x - 10

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3 years ago

0.85 in fractional notation

artcher [175]

If you would like to write 0.85 in a fractional notation, you can do this using the following steps:

0.85 = 85/100 which simplifies to 17/20 <span>(the common factor of 85 and 100 is 5, so you can divide both numbers by 5).

The correct result would be 17/20.</span>

8 0

3 years ago

Read 2 more answers

A teacher keeps track of the number of students that participate at least three times in an optional study session each year. He

Anastasy [175]

Answer:

It is not C got it wrong.

Step-by-step explanation:

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2 years ago

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If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

2 years ago