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marin [14]
4 years ago
12

I need help to Simplify i31

Mathematics
1 answer:
polet [3.4K]4 years ago
6 0

Answer:1/1

Step-by-step explanation:

What is 31/31 Simplified? - 1/1 is the simplified fraction for 31/31. Simplify 31/31 to the simplest form.

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It's 28 3/4

Step-by-step explanation:

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The concentration of dichlorodiphenyltrichloroethane (DDT - an infamous pesticide) in Lake Michigan has been declining exponenti
mojhsa [17]

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.

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She now has 6 oz. if that’s what you’re asking. If not please clarify
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That would be -9  (because theres a +9y in first equation).
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