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notsponge [240]

3 years ago

15

Gisselle has 1/2 of a candy bar. She role off 1/2 of what she had. She then took the piece she broke off and split it into three

equal pieces. What fraction of a candy bar was each piece?

1 answer:

saul85 [17]3 years ago

4 0

Answer:

$\frac{1}{12}$

Step-by-step explanation:

Given:

Gisselle has 1/2 of a candy bar.

She role off 1/2 of what she had and split it into three equal pieces.

Question asked:

What fraction of a candy bar was each piece?

Solution:

Gisselle has = $\frac{1}{2} \ of \ a \ candy \ bar$

She role off = $\frac{1}{2} \ of \ what\ she \ had$

$=\frac{1}{2} \ of\ \frac{1}{2}\\=\frac{1}{2} \times \frac{1}{2}\\\\=\frac{1}{4}$

As she split it into three equal pieces, we will divided it by 3, we get,

$\frac{1}{4} \div3\\\frac{1}{4}\times\frac{1}{3}$

$\frac{1}{12}$

Therefore, fraction of a candy bar of each piece is $\frac{1}{12}$

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Answer:

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Step-by-step explanation:

1) The function of the box is:

$S = 2\cdot (w - 2\cdot x)\cdot x + 2\cdot (l-2\cdot x)\cdot x +(w-2\cdot x)\cdot (l-2\cdot x)$

$S = 2\cdot w\cdot x - 4\cdot x^{2} + 2\cdot l\cdot x - 4\cdot x^{2} + w\cdot l -2\cdot (l + w)\cdot x + l\cdot w$

$S = 2\cdot (w+l)\cdot x - 8\cdpt x^{2} + 2\cdot w \cdot l - 2\cdot (l+w)\cdot x$

$S = 2\cdot w\cdot l - 8\cdot x^{2}$

2) The maximum cutout is:

$2\cdot w \cdot l - 8\cdot x^{2} = 0$

$w\cdot l - 4\cdot x^{2} = 0$

$4\cdot x^{2} = w\cdot l$

$x = \frac{\sqrt{w\cdot l}}{2}$

The domain of S is $0 \leq x \leq \frac{\sqrt{w\cdot l}}{2}$ . The range of S is $0 \leq S \leq 2\cdot w \cdot l$

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$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1\,in)^{2}$

$S = 176\,in^{2}$

4) The size is found by solving the following second-order polynomial:

$20\,in^{2} = 2 \cdot (8\,in)\cdot (11.5\,in)-8\cdot x^{2}$

$20\,in^{2} = 184\,in^{2} - 8\cdot x^{2}$

$8\cdot x^{2} - 164\,in^{2} = 0$

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$V = [w\cdot l -2\cdot (w+l)\cdot x + 4\cdot x^{2}]\cdot x$

$V = w\cdot l \cdot x - 2\cdot (w+l)\cdot x^{2} + 4\cdot x^{3}$

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$V = (92\,in^{2})\cdot x - (39\,in)\cdot x^{2} + 4\cdot x^{3}$

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$V' = 92\,in^{2} - (78\,in)\cdot x + 12\cdot x^{2}$

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$12\cdot x^{2}-(78\,in)\cdot x + 92\,in^{2} = 0$

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$x \approx 1.548\,in$

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$S = 2\cdot (8\,in)\cdot (11.5\,in)-8\cdot (1.548\,in)^{2}$

$S = 164.830\,in^{2}$

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Given:

Sides are 6.03 units and 7.96 units

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