Answer:
A)
y^2-6y = 0
or, y(y-6) = 0
or, y = 0 or y = 6
B)
n^2+5n+7 = 7
or, n^2+5n+7-7 = 7-7 ( Subtracting 7 from both sides)
or, n^2+5n = 0
or, n(n+5) = 0
or, n=0 or n= -5
C)
2t^2-14t+3 = 3
or, 2t^2-14t = 0
or, 2t(t-7) = 0
or, t=0 or t=7
D)
1/3x^2+3x-4 = -4
or, 1/3x^2+3x = 0
or, 1/3x(x+9) = 0
or, x=0 or x= -9
E)
Zero is a common solution to each of the equations. This is because each of the equations had a variable outside the parenthesis with an operation of multiplication.
THANK YOU FOR READING.
A cube has equal edges, so 2*2*2 or 2^3 (2 cubed) v=8^3
You can set up a system of equations for this problem. x= number of coach tickets and y = number of first class tickets.
$210x + $1200y = $10,230 (cost of coach ticket plus cost of first class tickets is total budget)
x + y = 11 (number of coach tickets plus number of first class tickets is total number of people)
Solve the second equation for y to get y = 11 - x, then plug that into the first equation and solve for x:
$210x + $1200(11 - x) = $10,230
$210x + $13,200 - $1200x = $10,230
-$990x + $13,200 = $10,230
-$990x = $2,970
x = 3
Sarah bought x = 3 coach tickets. Plug that into the second equation and solve for y:
3 + y = 11
y = 8
Sarah bought y = 8 first class tickets.
Answer:
2) no
3) yes
Step-by-step explanation:
i think. sorry if im wrong
