If the sides of the pentagon are the same length, it is probably a regular pentagon. If the angles are all the same within the shape, it is definitely a regular pentagon.
Answer:
q = 15
Step-by-step explanation:
Given
f(x) = x² + px + q , then
f(3) = 3² + 3p + q = 6 , that is
9 + 3p + q = 6 ( subtract 9 from both sides )
3p + q = - 3 → (1)
---------------------------------------
f'(x) = 2x + p , then
f'(3) = 2(3) + p = 0, that is
6 + p = 0 ( subtract 6 from both sides )
p = - 6
Substitute p = - 6 into (1)
3(- 6) + q = - 3
- 18 + q = - 3 ( add 18 to both sides )
q = 15
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
