How many pennies could you have if: When you break the pennies into groups of 2, you have 1 penny left over, AND when you break the pennies into groups of 3, you have 1 penny left over, AND when you break the pennies into groups of 5, you have 1 penny left over, AND when you break the pennies into groups of 7, you have NO pennies left over?

91

Step-by-step explanation:

"When you break the pennies into groups of 2, you have 1 penny left over"

You have an odd number of pennies.

"when you break the pennies into groups of 3, you have 1 penny left over"

The number of pennies is one more than a multiple of 3.

"when you break the pennies into groups of 5, you have 1 penny left over"

A number that is divisible by 5 ends in 0 or 5. You have one more penny than that, so the number will end in 1 or 6. We already know the number is odd, so it cannot end in 6. The number of pennies must end in 1.

"when you break the pennies into groups of 7, you have NO pennies left over"

The number is a multiple of 7.

We need a multiple of 7 that ends in 1.

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

You could have 91.

step-by-step explanation:

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1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.