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4 years ago

SOMEONE HELP!!! AHHHH!!!

Mathematics

1 answer:

asambeis [7]4 years ago

8 0

Hey there! :)

Answer:

The first option:

Y-intercept: 9

Zeros: 1, -1, 3, -3

Step-by-step explanation:

The Y-intercept is where the graph intersects with the y-axis. On this graph, we can see that the graph intersects at y = 9.

The zeros of the graph are where the graph intersects the x-axis. When looking at the graph, the x-axis is intersected at x = -3, x = -1, x = 1, and x = 3.

Therefore, A, or the first option, is the correct answer.

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A chemical substance has a decay rate of 6.9% per day. The rate of change of an amount N of the chemical is given by the equatio

GaryK [48]

Answer:

a) $N=N_0e^{-0.069t}$

b) $N=696.9$ grams

c) $t=10$ days

Step-by-step explanation:

We are going to use separation of variables to solve.

Get all your t's to one side and your N's to opposing side.

$\frac{dN}{dt}=-0.069N$

Multiply both sides by $dt$ :

$dN=-0.069N dt$

Divided both sides by $N$ :

$\frac{dN}{N}=-0.069 dt$

Integrate both sides:

$\ln|N|=-0.069t+C$

The equivalent exponential form is:

$e^{-0.069t+C}=N$

Using law of exponents you can write this as:

$e^{-0.069t}e^C=N$

$e^C$ is just a positive constant that I'm going to replace with K:

$e^{-0.069t}K=N$

Applying the symmetric property of equality:

$N=e^{-0.069t}K$

Applying the commutative property of multiplication:

$N=Ke^{-0.069t}$

K actually represents the initial amount of chemical substance since when plugging in 0 for t you get K for N, like so:

$N=Ke^{-0.069 \cdot 0}$

$N=Ke^{0}$

$N=K(1)$

$N=K$

We are given at time 0 the amount of chemical substance,N, is K. They want us to represent this value with $N_0$ instead. So the exponential equation is:

$N=N_0e^{-0.069t}$

We are given $N_0=800$ at $t=0$ .

We are asked to find how much of the chemical substance, N, remains after 2 days. So we replace t with 2 in $N=800e^{-0.069t}$ :

$N=800e^{-0.069 \cdot 2}$

Put into calculator:

$N=696.9$ (this was rounded to the nearest tenths)

The last part is asking for how many days will it take a initial 800 grams to go down to half of 800 grams.

We need to see the following equation:

$\frac{1}{2}(800)=800e^{-0.069t}$

$400=800e^{-0.069t}$

Divide both sides by 800:

$\frac{400}{800}=e^{-0.069t}$

Reduce the fraction:

$\frac{1}{2}=e^{-0.069t}$

Convert to logarithmic form:

$\ln(\frac{1}{2})=-0.069t$

Divide both sides by -0.069:

$\frac{\ln(\frac{1}{2})}{-0.069}=t$

Input into calculator:

$10.0=t$

$t=10.0$

$t=10$

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