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P(3)=17 indicates that when kerosene is passed through 3 feet of clay then 17ppm of pollutants are removed.

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Work was not written out properly, but answers are boxed

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Julia, Lola, Luc, Taylor

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With Bodhi, Alex S, and Alex J

We set f(x) as equal to 2, and then solved for x.
5x^2=f(x)
f(x)=2
5x^2=(fx)
x= (2/5)^-2

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Alex S. Alex J. and Bodhi
the zeroes of h are (1)
H(0)=-4
h(h(h(0)))=h(h(-4)=h(7)=-5

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I don’t know if my notation is correct

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This is a rough sketch of the graph for problem 2 that really helps in when trying to figure out all the side problems for this question.

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Evaluate h(0)
h(0) = -4

Solve h(x) = -4
x = 0, x = 3

Zeroes of h: 1

Find h(h(h(0)))
h(h(-4)
h(7) = -5

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With Isabella, Josie, and Felix:

So using the set we were able to find that h(0)=-4 Then the question asks to evaluate h(h(h(0))).

h(h(h(0)))=h(h(-4))
h(h(-4))=h(7)
h(7)=-5

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h(0)
zeros of h=-4
solve h(x)=-4
x=0, x=3
h(h(h(0)))
h(h(-4))
h(7)=-5

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h(0)=-4,h(-4)=7,h(7)=-5 This isnt correct notation just wanted to explain our thinking

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h(0)=-4,h(-4)=7,h(7)=-5 This isnt correct notation just wanted to explain our thinking

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Alex J. Alex S. and Bodhi
g(x)=3x^2+7x
find the zeros of g
g(0)=3(0)^2+7(0)=0+0=0

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for problem 3, do you need to wright solve before finding x?

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for problem 3, do you need to wright solve before finding x?

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g(x)=3x^2+7x

Solve:
0=3x^2+7x
0=3x(x+7/3)
3x=0, x+7/3=0
x=0, x=-7/3

x=0=0 is redundant.

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For question 3 pt 2, can you solve by completing the square? Is this necessary?