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.

your answer is the last one

step-by-step explanation:

divide the length of a'c' by the length of ac to get a scale factor of two

& then reflect vertically (over the y-axis)

so your answer is dilation by a scale factor of 2 followed by reflection over the y-axis

no this is a nonlinear negative