You have quarters and dimes that total $2.80. Your friend says it is possible that the number of quarters is 8 more than the number of dimes. Is your friend correct? Let d represent the number of dimes An equation that models this situation is =2.80
(I’m so stupid help)

Step-by-step explanation:

d = number of dimes

2.80 = 0.1d + 0.25 (d + 8 )

2.80 = 0.1d + 0.25d + 2            Combine like terms

2.80 = 0.35d + 2                       Subtract 2 from both sides

0.80 = 0.35d                             Divide both sides by 0.35

d ≈ 2.29

No, it's not possible to have more quarters than dimes when the total is $2.80 because it's not possible to have part of a dime.

a

step-by-step explanation:

answer is 8x+24

area of vanessa's patio = length × width

given, length = (3x² + 5x + 10)

width = (x² - 3x - 1)

area of vanessa's patio = (3x² + 5x + 10)(x² - 3x - 1)

= 3x²(x² - 3x - 1) + 5x(x² - 3x - 1) + 10(x² - 3x - 1)

= 3x⁴ - 9x³ - 3x² + 5x³ - 15x² - 5x + 10x² - 30x - 10

= 3x⁴ - 4x³ - 8x² - 35x - 10

hence, expression of vanessa's patio = (3x⁴ - 4x³ - 8x² - 35x - 10)