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Ksju [112]
3 years ago
13

1 pointHow does the graph of f(x) = 4(x – 3)2 – 2 compare to the graph of g(x) = x??​

Mathematics
1 answer:
Alex777 [14]3 years ago
4 0

Answer:

Graph f(x)=4(x-3)2-2 is simplified to f(x)=4x-12 which will not be as steep as g(x)=x because the slope is smaller. The y-intercept of F(x)=4(x-3)2-2 is -12 which is smaller than the graph g(x)=x, which is the orgin (0,0).

Step-by-step explanation:

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Answer: A) y=13e^{-0.1359t}

              B) H = 5.10

              C) Yes

Step-by-step explanation: <u>Exponential</u> <u>Decay</u> <u>function</u> is a model that describes the reducing of an amount by a constant rate over time. Generally, it is written in the form: y(t)=Ce^{rt}

A) C is initial quantity, in this case, the initial concentration of DDT. To determine r, using the data given:

y(t)=Ce^{rt}

2.22=13e^{13r}

e^{13r}=0.1708

Using a natural logarithm property called <em>power rule:</em>

13r=ln(0.1708)

r=\frac{ln(0.1708)}{13}

r=-0.1359

The decay function for concentration of DDT through the years is y(t)=13e^{-0.1359t}

B) The value of H is calculated by y=C(0.5)^{\frac{t}{H} }

2.22=13(0.5)^{\frac{13}{H} }

(0.5)^{\frac{13}{H} }=0.1708

Again, using power rule for logarithm:

\frac{13}{H} log(0.5)=log(0.1708)

\frac{13}{H} =\frac{log(0.1708)}{log(0.5)}

\frac{13}{H} =2.55

H = 5.10

Constant H in the half-life formula is H=5.10

C) Using model y(t)=13e^{-0.1359t} to determine concentration of DDT in 1995:

y(24)=13e^{-0.1359.24}

y(24) = 0.5

By 1995, the concentration of DDT is 0.5 ppm, so using this model is possible to reduce such amount and more of DDT.

8 0
3 years ago
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