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Sav [38]
1 year ago
6

There is a part c I will have to send it in the text there is not enough room

Mathematics
1 answer:
NemiM [27]1 year ago
5 0

we get that the first bowl counts for 2 inches and then 4 bowls count 3 inches. So the equation is

y=2+\frac{3}{4}(x-1)=2-\frac{3}{4}+\frac{3}{4}x=\frac{5}{4}+\frac{3}{4}x

b) x represents the number of bowls starting in 1

y represents the height of the tower of bowls

c)

y=\frac{5}{4}+\frac{3}{4}\cdot10=\frac{5}{4}+\frac{30}{4}=\frac{35}{4}

so the height is 35/4 inches

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Answer:

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Artist 52 [7]
Check the picture below.  Recall, is an open-top box, so, the top is not part of the surface area, of the 300 cm².  Also, recall, the base is a square, thus, length = width = x.

\bf \textit{volume of a rectangular prism}\\\\&#10;V=lwh\quad &#10;\begin{cases}&#10;l = length\\&#10;w=width\\&#10;h=height\\&#10;-----\\&#10;w=l=x&#10;\end{cases}\implies V=xxh\implies \boxed{V=x^2h}\\\\&#10;-------------------------------\\\\&#10;\textit{surface area}\\\\&#10;S=4xh+x^2\implies 300=4xh+x^2\implies \cfrac{300-x^2}{4x}=h&#10;\\\\\\&#10;\boxed{\cfrac{75}{x}-\cfrac{x}{4}=h}\\\\&#10;-------------------------------\\\\&#10;V=x^2\left( \cfrac{75}{x}-\cfrac{x}{4} \right)\implies V(x)=75x-\cfrac{1}{4}x^3

so.. that'd be the V(x) for such box, now, where is the maximum point at?

\bf V(x)=75x-\cfrac{1}{4}x^3\implies \cfrac{dV}{dx}=75-\cfrac{3}{4}x^2\implies 0=75-\cfrac{3}{4}x^2&#10;\\\\\\&#10;\cfrac{3}{4}x^2=75\implies 3x^2=300\implies x^2=\cfrac{300}{3}\implies x^2=100&#10;\\\\\\&#10;x=\pm10\impliedby \textit{is a length unit, so we can dismiss -10}\qquad \boxed{x=10}

now, let's check if it's a maximum point at 10, by doing a first-derivative test on it.  Check the second picture below.

so, the volume will then be at   \bf V(10)=75(10)-\cfrac{1}{4}(10)^3\implies V(10)=500 \ cm^3

6 0
3 years ago
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Nutka1998 [239]

Answer:

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Hope this helped! :)

7 0
2 years ago
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