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tia_tia [17]
2 years ago
11

Seth has 15 cans of regular soda and 20 cans of diet soda. He wants to create some identical refreshment tables that will operat

e during the high school football game. He also doesn't want to have any sodas left over. What is the greatest number of refreshment tables that Seth can stock? How many regular sodas will be at each table? How many diet sodas will be at each table?​
Mathematics
1 answer:
arsen [322]2 years ago
8 0
The greatest number is 5. 3 cans of regular and 4 cans of diet. The most 15 can be divided by similarly to 20 is by 5.
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Step-by-step explanation:

you put the y's on the top and the x's on the bottom and subtract. what you get is your slope.

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5/6(2x+12)-8=4-(x+1)
alekssr [168]

For this case we have the following equation:

\frac {5} {6} (2x + 12) -8 = 4- (x + 1)

We apply distributive property on the left side of the equation:

\frac {5 * 2} {6} x + \frac {5 * 12} {6} -8 = 4- (x + 1)\\\frac {10} {6} x + \frac {60} {6} -8 = 4- (x + 1)

We simplify the left side of the equation:

\frac {5} {3} x + 10-8 = 4- (x + 1)

Different signs are subtracted and the major sign is placed.

\frac {5} {3} x + 2 = 4- (x + 1)

On the right side we must take into account that:

- * + = -

So:

\frac {5} {3} x + 2 = 4-x-1\\\frac {5} {3} x + 2 = 3-x

We add x to both sides of the equation:

\frac {5} {3} x + x + 2 = 3\\\frac {5 * 1 + 3 * 1} {3} x + 2 = 3\\\frac {5 + 3} {3} x + 2 = 3\\\frac {8} {3} x + 2 = 3

We subtract 2 from both sides of the equation:

\frac {8} {3} x = 3-2\\\frac {8} {3} x = 1

We multiply by 3 on both sides of the equation:

8x = 3

We divide by 8 on both sides of the equation:

x = \frac {3} {8}

Thus, the solution of the equation is:

x = \frac {3} {8}

Answer:

x = \frac {3} {8}

4 0
4 years ago
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