-6(-6x + 1) plz answer
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Number of dimes were 81 and number of quarters were 47
<em><u>Solution:</u></em>
Let "d" be the number of dimes
Let "q" be the number of quarters
We know that,
value of 1 dime = $ 0.10
value of 1 quarter = $ 0.25
<em><u>Given that There are 128 coins in all</u></em>
number of dimes + number of quarters = 128
d + q = 128 ------ eqn 1
<em><u>Also given that collection of dimes and quarters is worth $19.85</u></em>
number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85
0.1d + 0.25q = 19.85 -------- eqn 2
<em><u>Let us solve eqn 1 and eqn 2</u></em>
From eqn 1,
d = 128 - q -------- eqn 3
<em><u>Substitute eqn 3 in eqn 2</u></em>
0.1(128 - q) + 0.25q = 19.85
12.8 - 0.1q + 0.25q = 19.85
12.8 + 0.15q = 19.85
0.15q = 7.05
<h3>q = 47</h3>Therefore from eqn 3,
d = 128 - q
d = 128 - 47
<h3>d = 81</h3>Thus number of dimes were 81 and number of quarters were 47
Sense -4 and +2 would be 6 digits away from each other, we would then have to figure out, what would 4 times x - 2 would equal "x".
We would do, 4×2 = 8.
8 - 2 = 6.
Sense 6 is the amount of how much -4 and +2 would be from each other, this would then mean that x would then equal 8.
We would do, 4×2 = 8.
8 - 2 = 6.
Sense 6 is the amount of how much -4 and +2 would be from each other, this would then mean that x would then equal 8.