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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?
Mathematics
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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First set up the equation.

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Divide both sides by 4.

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A garden table and a bench cost $781 combined. The garden table costs $69 less than the bench. What is the cost of the bench?
Arlecino [84]

Answer:

$425

Step-by-step explanation:

x = bench

x - 69 = garden table

x + (x - 69)=781

x + x - 69 =781

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Which graph represents the following system of inequalities.
Marta_Voda [28]

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After heating up in a teapot, a cup of hot water is poured at a temperature of 204°F.
laila [671]

Answer:

  • Question 1: 0.081 min⁻¹
  • Question 2: 168ºF

Explanation:

The missing equation is:

             T=T_a+(T_0-T_a)e^{-kt}

You are given:

  • Ta = the temperature surrounding the object = 73ºF
  • T₀ = the initial temperature of the object = 204ºF
  • t = 2.5 min
  • T after 2.5 min = 180ºF

Question 1: Find the the value of k.

Substitute the data into the equation and solve for k:

          180\º=73\º+(204\º-73\º)e^{-k\cdot 2.5min}\\ \\ 107\º=131\ºe^{-k\cdot 2.5min}\\\\-k\cdot 2.5min=\ln (107/131)\\\\k=0.081min^{-1}

Question 2: Find the Fahrenheit temperature of the cup of water, to the nearest degree, after 4 minutes.

Subsititute into the formula and compute:

            T=73\ºF+(204\ºF-73\ºF)e^{-0.081min^{-1}\times 4min}\\ \\ T=73\ºF+(131\ºF)(0.72325024)\\\\T=168\ºF

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