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kenny6666 [7]
2 years ago
14

You have a 3-gallon and a 5-gallon jug that you can fill from a fountain of water. The problem is to fill one of the jugs with e

xactly 4 gallons of water. How do you do it?
Mathematics
1 answer:
Nataly [62]2 years ago
7 0

Answer:

See below

Step-by-step explanation:

Fill the 5   and pour into the 3    ....this will leave 2 gal in the 5 bucket

  empty the 3 bucket and put that 2 gallon in to it.....

      then fill the 5 bucket again......pour 1 gallon into the 3 gallon bucket to fill it....then there will be 4 gallons left in the 5 bucket

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3 years ago
Which is the approximate solution to the system y = 0.5x + 3.5 and y = − 2/3 x + 1/3 shown on the graph? (–2.7, 2.1) (–2.1, 2.7)
AlekseyPX

Answer:

The approximate solution to the system is (-2.7, 2.1).

Step-by-step explanation:

To solve the system of equations \begin{bmatrix}y=0.5x+3.5\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix} you must:

\mathrm{Rationalize\:equations}\\\\\begin{bmatrix}y=\left(\frac{1}{2}\right)x+\left(\frac{7}{2}\right)\\ y=-\frac{2}{3}x+\frac{1}{3}\end{bmatrix}

\mathrm{Subsititute\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\begin{bmatrix}-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\end{bmatrix}

\mathrm{Isolate}\:x\:\mathrm{for}\:-\frac{2}{3}x+\frac{1}{3}=\frac{1}{2}x+\frac{7}{2}\\\\-\frac{2}{3}x=\frac{19}{6}+\frac{1}{2}x\\\\-\frac{7}{6}x=\frac{19}{6}\\\\6\left(-\frac{7}{6}x\right)=\frac{19\cdot \:6}{6}\\\\-7x=19\\\\x=-\frac{19}{7}\approx-2.7

\mathrm{For\:}y=-\frac{2}{3}x+\frac{1}{3}\\\\\mathrm{Subsititute\:}x=-\frac{19}{7}\\\\y=-\frac{2}{3}\left(-\frac{19}{7}\right)+\frac{1}{3}\\\\y=\frac{15}{7}\approx 2.1

The approximate solutions to the system of equations are:

x=-2.7 ,\:y=2.1

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