In one day, a toy store sold 15 toy cars and 20 yo-yos for a total of $190. the next day, the toy store sold 3 toy cars and 12 yo-yos for a total of $102. what is the cost of one toy car?
$1 for one car
$2 for one car
$4 for one car
$8 for one car

Option B is the correct answer.

Step-by-step explanation:

Let the cost of toy car be p and yo yo be q.

We have in one day, a toy store sold 15 toy cars and 20 yo-yos for a total of $190

                15 p + 20 q = 190 eqn1

The next day, the toy store sold 3 toy cars and 12 yo-yos for a total of $102

                3 p + 12 q = 102 eqn2

eqn2 x 5 gives

                 15 p + 60 q = 510 eqn3

eqn3 - eqn1 gives

                 15 p + 60 q - (  15 p + 20 q ) = 510 - 190

                 40 q = 320

                       q = 8 $

Substituting in eqn1

                 15 p + 20 x 8 = 190

                    p = 2

Price of toy car = 2 $

Option B is the correct answer.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18

the answer is 96 degrees

step-by-step explanation:

you have to realize that a line equals 180 degrees