Help with question only 1

17 adults, 13  students and 90 children

Step-by-step explanation:

in order to answer his question we have to create two equations

we first need to give each group of people a variable:

adults = x

student = y

children = z

first equation is for the total price

5x + 2y + 0.10z = 120

second equation is for the total number of people :

x + y+ z = 120

- in this case substitution and elimination method can not work as there are three unknown variable not two

in order for the total amount from the ticket sales to be exactly $120 , then the number of children have to be a multiple of 10 ( eg 0.10 * 10 = 10)

therefore they will be no decimal point

through the use of probability and trial and error

number of   number of   number of    total cost

    children       adults        students      of tickets

       90           1          29           $ 72

       90           3          27           $ 78

       90           5          25           $ 84

       90           7          23           $ 90

       90           9          21           $ 96

       90          11          19           $102

       90          13          17           $108

       90          15          15           $114

       90          17          13           $120

we can test this put in the equation :

5x + 2y + 0.10z = 120

5 (17) + 2 (13) + 0.10 (90) = 120

also the other equation can be used to check if it is correct :

x + y+ z = 120

17 + 13 + 90 = 120

hence this answer is CORRECT

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

yes because there is exactly one y value for every x value

for function there can't be more than one value for every x value