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ycow [4]
3 years ago
8

Select correct answer pls^^

Mathematics
2 answers:
alina1380 [7]3 years ago
4 0

It takes 48 hours if 12 people built the same wall.

Please see the attached picture for full solution

Hope it helps

Good luck on your assignment

Scilla [17]3 years ago
3 0

Answer:

3 x 12 x 129

Step-by-step explanation:

You can get your answer

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The table below shows the coordinates of a figure that was transformed. Which is a correct description of the transformation?
Hitman42 [59]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
The question in the image.
GREYUIT [131]

Answer:

500 boxes

Step-by-step explanation:

<em>hey there,</em>

<em />

< Since "p" stands for profits, change "p" to 0 (since the question says 0 profits).

Here is how your equation would look like:

0 = -n^2 + 300n + 100000

Pretend like "n" is "x" since that is the variable we are trying to find here.

Solve for "n". Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

"n" actually ends up equaling 500, -200.

Obviously, negative can't be the answer because you can't have a negative amount of boxes. So 500 boxes would be your answer. 200 CAN'T be your answer <em>either </em>because it is a negative!! >

<u>Hope this helped! Feel free to ask anything else.</u>

6 0
3 years ago
WILL MARK BRAINLIEST. Can someone help me wit des 2 questions?
Mnenie [13.5K]

Answer:

Zero is a number that can be equal to its opposite.

So, the given equation has solution for which LHS=RHS=0.

Step-by-step explanation:

The answer to the equation is 2

7 0
3 years ago
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
If you're good at trig pls help meeeeee and show full working out pls
saveliy_v [14]

Answer:

19 km

Step by step explanation:

6 0
3 years ago
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