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tigry1 [53]
3 years ago
13

Show your work!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Answer:

Here is my workk

Step-by-step explanation:

WORKKKKK

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What is the question
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4 years ago
Ed wants to buy a new video game system which costs $250. he has
BARSIC [14]

The required equation is <u>135 + 9x > 250</u>.

The number of lawns Ed must mow is assumed to be x.

The amount Ed charges for each lawn he mows is $9.

Thus, the total amount Ed earns by mowing x lawns = $9x.

The savings which Ed has is $135.

Thus, the total amount Ed will have to spend can be written as the expression, $(135 + 9x).

The cost of the video game is given to be $250.

We are asked to write an equation, that can be used to find the number of lawns Ed mow, that is x so that the amount Ed has will be more than the amount he needs to buy the video game.

This can be shown as the equation:

Total amount Ed has > Cost of the video game,

or, 135 + 9x > 250.

Thus, the required equation is <u>135 + 9x > 250</u>.

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2 years ago
There are 24 people using the gym the ratio of men to women is 2:1 how many men are using the gym?
elixir [45]
I find it easier if you draw out people(not perfectly just make a circle or something) and then circle every 2 people to count for the men. There are 16 men in the gym.
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3 years ago
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A. less than three batteries were used in five hours.

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3 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
3 years ago
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