Bob has $3.69 in pennies, dimes, and quarters. The number of pennies is three more than the number of dimes. The number of quarters is twice the number of dimes. How many of each coin does he have?

9 pennies, 6 dimes, 12 quarters.

Step-by-step explanation:

Pennies = $0.01

Dimes = $0.10

Quarters = $0.25

Let x be the number of dimes. Then (x + 3) is the number of pennies and 2x is the number of quarters.

We have x($0.10) + (x + 3)($0.01) + 2x($0.25) = $3.69

x($0.10) + (x + 3)($0.01) + 2x($0.25) = $3.69

$0.1x + $0.01x + $0.03 + $0.5x = $3.69

$0.61x = $3.66

x = 6.

Therefore Bob has 9 pennies, 6 dimes and 12 quarters.

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81