Answer:
10 quarters = $2.50
10 nickels = $0.50
that leaves $0.20 for other coins (dimes / pennies)
Step-by-step explanation:
First, suppose she has only quarters and nickels and no other coins. Then if C is the identical number of coins of each type, then 5C + 25C = 320, so 30C = 320 and 3C = 32, but there is no integer solution to this. So she must have at least one other type of coin.
Assume she has only quarters, nickels, and dimes. Then if D is the number of dimes, 5C + 25C + 10D = 320, which means 30C + 10D = 320, or 3C + D = 32. The smallest D can be is 2, leaving 3C = 30 and thus C = 10. So in this scenario she would have 10 quarters, 10 nickels, and two dimes to make $2.50 + $0.50 + $0.20 = $3.20.
This has to be the highest number, because if she had 11 quarters and 11 nickels, that alone would add up to 11(0.25) + 11(0.05) = $3.30, which would already be too much.
3x+22.5 = 69.75
-22.5 -22.5
3x = 47.25
/3 /3
x = 15.75
The price of each shirt before tax is $15.75
Notice that
11/12 = 1/6 + 3/4
so that
tan(11π/12) = tan(π/6 + 3π/4)
Then recalling that
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
⇒ tan(x + y) = (tan(x) + tan(y))/(1 - tan(x) tan(y))
it follows that
tan(11π/12) = (tan(π/6) + tan(3π/4))/(1 - tan(π/6) tan(3π/4))
tan(11π/12) = (1/√3 - 1)/(1 + 1/√3)
tan(11π/12) = (1 - √3)/(√3 + 1)
tan(11π/12) = - (√3 - 1)²/((√3 + 1) (√3 - 1))
tan(11π/12) = - (4 - 2√3)/2
tan(11π/12) = - (2 - √3) … … … [A]
Answer:
the answer is 1
Step-by-step explanation:
69 goes into 69, 1 time
