Acollection of dimes and quarters is worth $9.55. if the quarters were dimes and the dimes were quarters, the total value would be 7.60. find the number of each coin.

Number of dimes are 18 and number of quarters are 31

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

A collection of dimes and quarters is worth $9.55

value of 1 dime x number of dimes + value of 1 dime x number of quarters = 9.55

0.10d + 0.25q = 9.55 eqn 1

If the quarters were dimes and the dimes were quarters, the total value would be 7.60

quarters were dimes means , q = d

dimes were quarters means d = q

0.25d + 0.10q = 7.60 eqn 2

Let us solve eqn 1 and eqn 2 to find "d" and "q"

Multiply eqn 1 by 2.5

0.25d + 0.625q = 23.875 eqn 3

Subtract eqn 2 from eqn 3

0.25d + 0.625q = 23.875

0.25d + 0.10q = 7.60

( - )

0.525q = 16.275

q = 31

Substitute q = 31 in eqn 1

0.10d + 0.25q = 9.55

0.10d + 0.25(31) = 9.55

0.10d + 7.75 = 9.55

0.10d = 1.8

d = 18

Thus dimes are 18 and number of quarters are 31

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256π units²   (answer b)

step-by-step explanation:

the formula for the surface area of a sphere of radius r

is             a = 4πr²

and so, if r = 8 units, then the surface area is:

a = 4π(8 units)² = 256π units²   (answer b)

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation: