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Burka [1]
3 years ago
10

Consider a collection of pennies with the following constraints: When the pennies are put in groups of 2 there is one penny left

over. When they are put in groups of three, five and six there is also one penny left over. But when they are put in groups of seven there are no pennies left over. How many pennies could there be?
Mathematics
1 answer:
Kipish [7]3 years ago
6 0

Answer:

91 pennies.

Step-by-step explanation:

It is given that 5 types of groups of pennies: group of 2, group of 3, group of 5, group of 6, and group of 7. If the pennies are arranged in the groups of 7, no pennies are left over. This means that the number of pennies has to be a multiple of 7 in order to satisfy this constraint. In the rest of the groups, there will always be 1 penny remaining without any remaining. This means that the number of pennies will yield the remainder of 1 if it is divided by 2, 3, 5, and 6. Possible multiples of 7:

0, 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98...

The number which satisfies the above conditions is 91. Since 91 divided by 7 is 13 and if 91 is divided by other numbers, the remainder will always be 1. Therefore, there are 91 pennies!!!

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Step-by-step explanation:

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Consider the following graph which represents the solutions to a system of inequalities. Which of the following systems of inequ
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Answer:

The system of inequalities is -2x + y > 0    -x + y ≥ 0

Step-by-step explanation:

The form of the equation of a line is y = m x + b, where

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  • b is the y-intercept

The line passes through the origin, then

  • The value of b = 0
  • The form of the equation is y = m x

Let us look at the graph to find the correct answer

∵ One of the lines are solid and the other is dashed

∵ The shaded area is over the two lines

∴ The signs of the two inequalities are ≥ and >

∵ The two lines pass through the origin

→ That means the y-intercepts are 0

∴ The form of the inequalities are y ≥ m_{1} x and y > m_{2}

There are only two answers that have these forms, so we must find the slope of each line

∵ m = \frac{y2-y1}{x2-x1} , (x1, y1) and (x2, y2) are two points on the line

∵ The solid line passes through points (0, 0) and (2, 2)

∴ m_{1} = \frac{2-0}{2-0} = \frac{2}{2} = 1

→ Substitute it in the form of the equation

∴ y ≥ 1(x)

∴ y ≥ x

→ Subtract x from both sides

∴ -x + y ≥ 0

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∴ m_{2} = \frac{2-0}{1-0} = \frac{2}{1} = 2

→ Substitute it in the form of the equation

∴ y > 2(x)

∴ y > 2x

→ Subtract 2x from both sides

∴ -2x + y > 0

The system of inequalities is -2x + y > 0    -x + y ≥ 0

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