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KATRIN_1 [288]
3 years ago
6

Maria made a cubical tank which can hold 42875 cubic meter of water. what is the height of the tank

Mathematics
1 answer:
ch4aika [34]3 years ago
5 0

Answer:

35 meters

Step-by-step explanation:

Since it's a cubical tank, its height, width and length are equal.

\sqrt[3]{42875}=35

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What sine function represents an amplitude of 1, a period of 2π, a horizontal shift of π, and a vertical shift of −4?
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\bf % template detailing&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\left. \qquad  \right.  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\left. \qquad  \right. if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\

\bf \bullet \textit{vertical shift by }{{  D}}\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\left. \qquad  \right. if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{function period or frequency}\\&#10;\left. \qquad  \right. \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;\left. \qquad  \right. \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)

with that template in mind, let's see

the period of 2π, well, for that, you have to do nothing, because that's sine original period

Amplitude of 1, well, that's also sine's original amplitude

horizontal shift/phase... ok, that means C/B = π, now, it doesn't say if to the left or the right, so I gather either is ok, so...let's do the right

you could use then, C = -π and B = 1, that gives you -π/1 or -π

vertical shift of -4, well, that simply means D = -4

f(θ) = sin(θ - π) - 4   <-- will do
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The vertex form needed is :

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Conversion will go like this:

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This is the resultant vertex form.

Thus the vertex form needed is :

-6(x-0.25)^2 +2.375

Learn more here:

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