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A car is moving east along a stralght, level road at a constant velocity. Which forces form an action/reaction force
pair?
o the force of the tires against the road and the force of the road on the tires
O the weight of the car and the force of the road pushing up on the car
the force of the engine on the car and the force due to friction on the car
o the normal force on the car and the force of the car pulling on Earth
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the force of the tires against the road and the force of the road on the tires

Step-by-step explanation:

i took the quiz

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3

c. 1,233 miles