Answer:
3X/20 (option a) of the pastries submitted by Rashid and Mikhail were brushed with butter
Step-by-step explanation:
Rashid pastries (R)
Mikhail pastries (M)
Rashid and Mikhail submitted a total of x pastries
R+M=x (I)
Rashid made 2/3 as many pastries as Mikhail
(2/3)*R=M (II)
Using II in I
R+(2/3)*R=x
(5/3)*R = x
R=(3/5)*x (III)
Using III in I
(3/5)*x+M=x
M=x-(3/5)*x
M=(2/5)*x (IV)
Mikhail filo dough (MF)
Mikhail shortcrust dough (MS)
Rashid filo dough (RF)
Rashid shortcrust dough (RS)
Mikhail used filo dough for all of his pastries
MF=M
MS=0
Rashid used shortcrust dough for all of his pastries
RS=R
RF=0
Filo dough (FD)
FD=RF+MF=0+MF=MF=M (V)
5/8 of the filo dough pastries were brushed with olive oil
pastries brushed with olive oil (OI)
(5/8)*FD=OI
Using V
(5/8)*M=OI
Using IV
(5/8)*(2/5)*x=OI
(1/4)*x=OI (VI)
pastries brushed with butter (B)
Pastries made out of filo dough are brushed with either olive oil or butter (but not both)
FD=OI+B
B=FD-OI
Using V and VI
B= M - (1/4)*x
Using IV
B = (2/5)*x - (1/4)*x
B= (3/20)*x
3X/20 (option a) of the pastries submitted by Rashid and Mikhail were brushed with butter
Looks like you going to summer school
No if its you sorry I don’t really mean it
Answer & Step-by-step explanation:
To find if they have a constant of proportionality of 12, use the following:
Divide y by the x value (x,y), and if the remaining equation is true, then that table has a constant of proportionality of 12.*
:Done
*Make sure you check all the values in a table. Sometimes only the first values will have k=12, while the others don't.
**The constant of proportionality is represented by <em>k</em>.
