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2 years ago

help. If 20 men can survive for 24 days on 15 cans of rations, how many cans will be needed for 16 men to survive for 36 days?

Mathematics

1 answer:

Semmy [17]2 years ago

4 0

Answer: 29 cans.

Step-by-step explanation:

20 can survive 24 days with 15 cans.

If X is the number of days that a man can survive with one can of rations.

so X is in the units can/day

then we have that:

24/15*X = 20

X = 20*15/24 = 12.5

This means that a man can live 12.5 days with a can of food.

then, for 16 men and 36 days we have:

(36/C)*12.5 = 16

C = (36/16)*12.5 = 28.1215

And we can not have a 0.1215 of a can, so we should round it up to 29 cans.

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Ahat [919]

Answer:

A) But if anyone thinks this they should go back to preschool

Step-by-step explanation:

On any other value of x you would take the numerators and denominators and multiply them across:

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But, this literally does not make since considering the fact that you cannot divide any value by zero. Try it, it won't work. If we plug these into a trusty dusty calculator, we get this:

$\frac{0}{3}=0\\\\\frac{6}{0}=DOMAINERROR$

0*DOMAINERROR= You can't do it.

There are other reasons why you can't divide by zero, I would suggest looking them up on YT and learning about them.

I would answer A, but the person who wrote this question just wasn't focusing when they wrote the question.

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3 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
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\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
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f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
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\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
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now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
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2 years ago

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Harlamova29_29 [7]

Answer:

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Step-by-step explanation:

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m = ∫2πr³∫sinθdθ]dr

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3 years ago

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Minchanka [31]

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