Let x = no. of 10 oz cups sold
Let y = no. of 14 oz cups sold
Let z = no. of 20 oz cups sold
:
Equation 1: total number of cups sold:
x + y + z = 24
:
Equation 2: amt of coffee consumed:
10x + 14y + 20z = 384
:
Equation 3: total revenue from cups sold
.95x + 1.15y + 1.50z = 30.60
:
Mult the 1st equation by 20 and subtract the 2nd equation from it:
20x + 20y + 20z = 480
10x + 14y + 20z = 384
------------------------ subtracting eliminates z
10x + 6y = 96; (eq 4)
Mult the 1st equation by 1.5 and subtract the 3rd equation from it:
1.5x + 1.5y + 1.5z = 36.00
.95x + 1.15y+ 1.5z = 30.60
---------------------------subtracting eliminates z again
.55x + .35y = 5.40; (eq 5)
Multiply eq 4 by .055 and subtract from eq 5:
.55x + .35y = 5.40
.55x + .33y = 5.28
--------------------eliminates x
0x + .02y = .12
y = .12/.02
y = 6 ea 14 oz cups sold
Substitute 6 for y for in eq 4
10x + 6(6) = 96
10x = 96 - 36
x = 60/10
x = 6 ea 10 oz cups
That would leave 12 ea 20 oz cups (24 - 6 - 6 = 12)
Check our solutions in eq 2:
10(6) + 14(6) + 20(12) =
60 + 84 + 240 = 384 oz
A lot steps, hope it made some sense! I hope this helps!! ;D
A number(p) plus six
six plus a number(p)
Answer:
3
Step-by-step explanation:
Answer:
B and E
Step-by-step explanation:
perpendicular sides mean they will equal 90 degrees... so all rectangles and squares are perpendicular- a 90 degree triangle also is perpendicular. The Equatorial and the Acute triangle do not have 90 degree angles
ANSWER
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EXPLANATION
We were given that,
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This implies that,
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We use the Pythagorean identity
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to get,

We were also given that,

This means that,

This is because,


This angles are in the first quadrant so we pick the positive values.




The correct answer is D