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.
idk if this is the right answer bc I had the same question with that same sentence and I got the correct answer so I decided to answer your question for u. Hope it helps.
Easy peasy just subsitute I(x) for the x in the h(x) so h(I(s))=-(2s+3)^2-4 distribute and simplify h(I(s))=-(4s^2+12s+9)-4 h(I(s))=-4s^2-12s-9-4 h(I(s))=-4s^2-12s-13
