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OlgaM077 [116]
3 years ago
5

A 24 inch board is cut into 3 pieces so that the second piece is twice as long as the first and the third is 3 times as long as

the first puece. If x represents the length of the first piece find the lengths of all three pieces
Mathematics
2 answers:
Lady_Fox [76]3 years ago
3 0

Answer:

The answer to your question is:  first piece = 4 in, second piece = 8 in, third piece = 12 in

Step-by-step explanation:

Data

total length = 24 inches

first piece = x in

second piece = 2x

third piece = 3x

Equation

               x + 2x + 3x = 24

               6x = 24

               x = 24/6

               x = 4               length of the first piece

second piece = 2(4) = 8 inches

third piece = 3(4) = 12 inches

yulyashka [42]3 years ago
3 0

Answer:

  1. 4 in
  2. 8 in
  3. 12 in

Step-by-step explanation:

The total length of the three pieces is ...

  x + 2x + 3x = 24 . . . . inches

  6x = 24 . . . . . . . . . . . collect terms

  x = 4 . . . . . the length in inches of the first piece

  2x = 8 . . . . the length in inches of the second piece

  3x = 12 . . . . the length in inches of the third piece

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The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i
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Answer:

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Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

S=\pi r\sqrt{r^2+h^2}

and volume,

V=\frac{1}{3}\pi r^2 h

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

f(r,h)=\frac{1}{3}\pi r^2 h

subject to,

g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)

We know for maximum volume r\neq 0. So let \lambda be the Lagranges multipliers be such that,

f_r=\lambda g_r

\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})

\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)

And,

f_h=\lambda g_h

\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}

\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)

Substitute (3) in (2) we get,

\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})

\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)

\implies h^2=2r^2

Substitute this value in (1) we get,

\pi r\sqrt{h^2+r^2}=8

\implies \pi r \sqrt{2r^2+r^2}=8

\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252

Then,

h=\sqrt{2}(1.21252)\equiv 1.71476

Hence largest volume,

V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114

3 0
3 years ago
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