A collection of nickels, dimes, and quarters consist of 19 coins with a total of $ 2.30 . If the number of dimes is equal to the number of nickels, find the number of each type of co

n = 7, d = 7, q = 5

Step-by-step explanation:

1 nickel = $0.05

1 dime = $0.10

1 quarter = $0.25

Let

Nickel = n

Dime = d

Quarter = q

n + d + q = 19 (1)

0.05n + 0.10d + 0.25q = 2.30 (2)

n = d

Substitute into (1)

n + n + q = 19

2n + q = 19

q = 19 - 2n

0.05n + 0.10n + 0.25q = 2.30

0.15n + 0.25q = 2.30

Substitute q = 19 - 2n

0.15n + 0.25(19 - 2n) = 2.30

0.15n + 4.75 - 0.5n = 2.30

0.15n - 0.5n = 2.30 - 4.75

-0.35n = -2.45

n = -2.45/-0.35

n = 7

Since n = d

then, d = 7

n + d + q = 19

7 + 7 + q = 19

14 + q = 19

q = 19 - 14

= 5

q = 5

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.