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KATRIN_1 [288]
2 years ago
7

You are filling bottles from 5 gallons of lemonade. How many bottles can you fill when each bottle is 3/8 of a gallon?

Mathematics
2 answers:
liraira [26]2 years ago
7 0

Answer:

13\frac{1}{3}  bottles.

Step-by-step explanation:

We have been given that you are filling bottles from 5 gallons of lemonade. We are asked to find the number of bottles each holding 3/8 of a gallon that can be filled from 5 gallons of lemonade.

To find the number of bottles, we will divide 5 by 3/8 as:

\text{Number of 3/8 gallon bottles}=5\div \frac{3}{8}

\text{Number of 3/8 gallon bottles}=\frac{5}{1}\div \frac{3}{8}

Convert to multiply problem by flipping the 2nd fraction:

\text{Number of 3/8 gallon bottles}=\frac{5}{1}\times  \frac{8}{3}

\text{Number of 3/8 gallon bottles}=\frac{5\times 8}{1\times 3}

\text{Number of 3/8 gallon bottles}=\frac{40}{3}

\text{Number of 3/8 gallon bottles}=13\frac{1}{3}

Therefore, 13\frac{1}{3} bottles of lemonade can be filled.

Crank2 years ago
5 0

5 gallons and each bottle is 3/8

8 times 5 equals 40

sooo =(total parts of the gallon being cut into eighths) 40/8

Now count by 3's to try to get as close as 40 :

3/8= 1 bottle

6/8= 2 bottles

9/8= 3 bottles

12/8= 4 bottles

15/8= 5 bottles

18/8= 6 bottles

21/8= 7 bottles

24/8= 8 bottles

27/8= 9 bottles

30/8=10 bottles

33/8= 11 bottles

36/8=12 bottles

39/8= 13 bottles

Answer: 13 bottles can be filled with 1/8 being left over

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Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6
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Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

Step-by-step explanation:

Given the simultaneous equations

y=9-x

y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations

y=9-x

-

\underline{y=2x^2+4x+6}

y-y=9-x-\left(2x^2+4x+6\right)

\mathrm{Refine}

x\left(2x+5\right)=3

\mathrm{Solve\:}\:x\left(2x+5\right)=3

2x^2+5x=3        ∵ \mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{Solve\:with\:the\:quadratic\:formula}

\mathrm{Quadratic\:Equation\:Formula:}

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\

x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

  =\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{4}

  =\frac{-5+7}{4}

  =\frac{2}{4}

  =\frac{1}{2}

Similarly,

x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{1}{2},\:x=-3

\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

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