Algebra

1) There are two pet stores in zacharys town. There are dogs, cats, and hamsters at the first pet store. The second store has dogs, rabbits, and fish. Write a set that represents all of the different pets stores in the town.

4) The cheerleading squad is having a bake sale. They want to bake at least 120 cookies. They made 12 cookies the first hour and 24 cookies the second hour.
a)Write an inequality that describes the scenario. Let Variable h represent the number of hours that must bake to meet or surpass the goal.

B) How many hours must they work to meet or surpass the goal?

6) John is at a car show. Beginning 2.5 miles away, a car traveling at a constant 45 miles per hour approaches and then passes John. The distance between John and the car can be represented by the equation d=/2.5-4.5t/. at what times is the car 0.5 miles from john?

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  1. if the stores are A and B, then

    S = {A,B}

    represents the stores. Now, if you want to represent the pets, then

    P1 = {{dog,cat,hamster},{dog,rabbit,fish}}

    Again, that is a set with two elements, each of which is a set describing the pets sold at that store.

    If all you want is to list all the pets sold in town, then that would be

    P2 = {dog,cat,hamster,rabbit,fish}

    For the cookies, you have given no indication of how many cookies they will bake in later hours. So far, in 3 hours, they have made 36 cookies, but I have no idea how many more hours it will take to do the remaining 84 cookies.

    For the car, we want to solve for t hours when

    |2.5-45t| = 0.5

    Now, recall that
    |x| = x if x >= 0
    |x| = -x if x < 0

    So, we have two cases:
    2.5-45t >= 0
    2.5-45t = 0.5
    45t = 2.0
    t = 2/45 = 4/90

    2.5 - 45t < 0
    -(2.5-45t) = 0.5
    -2.5 + 45t = 0.5
    45t = 3
    t = 1/15 = 6/90

    Now, you have to figure in that it takes the car .5/45 = 1/90 hours to pass John, so adjust your times accordingly.

    Or, you can consider that if the car passes John at t=0, it takes 1/90 hour to cover 1/2 mile, so the car is 1/2 mile away 1/90 hour before and after passing John. Since it takes 2.5/45 = 5/90 hours to get to john, it is 1/2 mile away at 4/90 and 6/90 hours from starting.

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    posted by Steve

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