<span>You forgot to tell us how much the tickets cost each.
So I just arbitrarily chose that the price of a student
ticket is $4 and the price of a non-student ticket is $6.
<span>Let the number of student tickets be x
Let the number of non-student tickets be y
Price Money
Type Number of from
of of EACH ALL
ticket tickets ticket tickets
-------------------------------------------
student x $4 $4x
non-student y $6 $6y
-------------------------------------------
TOTALS 48 ----- $220
The first equation comes from the second column.
x + y = 48
The second equation comes from the last column:
4x + 6y = 220
So we have this system of equations:
.
We solve by substitution. Solve the first equation for y:
x + y = 48
y = 48 - x
Substitute (48 - x) for y in 4x + 6y = 220
4x + 6(48 - x) = 220
4x + 288 - 6x = 220
-2x + 288 = 220
-2x = -68
x = 34 = the number of student tickets.
Substitute in y = 48 - x
y = 48 - (34)
y = 14 non-student tickets.
Checking: 34 student tickets brings in $136 and 14 non-students is $84
That's 48 tickets.
And indeed $136 + $84 = $220
</span>
</span>
Hello! When all of these numbers are added together, the sum is 222,126. Here’s the problem written vertically to show my work.
Answer:
#3
Step-by-step explanation:
Answer:
Step-by-step explanation:
d^2=h^2+w^2
a)
w^2=42^2-h^2, h=21
w^2=1764-441
w^2=1323
w=36.373
b)
36.373+2(2) by 21+2(2)
w=40.373, h=25
