Yes it’s right for this problem
2000, you just need to put three zeros at the end of the number.
Let a=price of adult ticket
let c=price of a child's ticket
start out by writing the following system of equations:
3a+4c=132
2a+3c=94
then, multiply the first equation by 2, and the second equation by 3 to get the following system of equations:
6a+8c=264
6a+9c=282
subtract the like terms to get the following equation:
-c=-18
divide both sides by -1 to get rid of the negative to get the price of a child's ticket to be $18. to find the price of an adult ticket, pick one of the original equations to substitute the 18 in for c to find a. for example:
2a+3c=94
2a+3(18)=94
2a+54=94
-54 -54
2a=40
2 2
a=20
or if you decide to use the other equation:
3a+4c=132
3a+4(18)=132
3a+72=132
-72 -72
3a=60
3 3
a=20
either way, you still get an adults ticket to be $20 and a child's ticket to be $18.
Step-by-step explanation:
4a² - 8ab + 4b²
= (2a)² - 2 * 2a * 2b + (2b) ²
= (2a - 2b) ²
Hope it helps :)
T(3)=3(3)-1
T(3)=9-1
T(3)=8
answer is 8
