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leonid [27]
3 years ago
11

Find the zeros, vertical asymptote, horizontal asymptote, slant asymptote, and graphing points.

Mathematics
1 answer:
sasho [114]3 years ago
6 0
First, simplify fraction
(x+4)/((x-2)(x+3))
good
set numerator to zero
x+4=0
x=-4

VA
set denom to zero
(x-2)(x+3)=0
VA's at x=-3 and x=2

HA
for p(x)/q(x)
degreee of p(x)<q(x)
HA=0
no slant asymtote
does it cross HA?
0=(x+4)/(x^2+x-6)
0=x+4
-4=x
it crosses at (-4,0)

to graph
plot point (-4,0)
draw assymtotes
y=0
x=2
x=-3

go from top left to middle botttom to top right
reverse 'L' on top left going through (-4,0)
upside down 'U' at middle
'L' at right top

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Find the value x.<br> Need help....thank you
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Step-by-step explanation:

It is parallel to the third side and has a length equal to half the length of the third side.

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3 years ago
Which polynomial is a quadratic trinomial?
il63 [147K]
<h3>Answer: Choice C</h3>

The largest exponent here is 4, so that makes it a quartic.

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3 years ago
Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species
kotegsom [21]

Answer:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

Step-by-step explanation:

The logistic equation is the following one:

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

In which P(t) is the size of the population after t years, K is the carrying capacity of the population, r is the decimal growth rate of the population and P(0) is the initial population of the lake.

In this problem, we have that:

Biologists stocked a lake with 80 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 2,000. This means that P(0) = 80, K = 2000.

The number of fish tripled in the first year. This means that P(1) = 3P(0) = 3(80) = 240.

Using the equation for P(1), that is, P(t) when t = 1, we find the value of r.

P(t) = \frac{KP(0)e^{rt}}{K + P(0)(e^{rt} - 1)}

240 = \frac{2000*80e^{r}}{2000 + 80(e^{r} - 1)}

280*(2000 + 80(e^{r} - 1)) = 160000e^{r}

280*(2000 + 80e^{r} - 80) = 160000e^{r}

280*(1920 + 80e^{r}) = 160000e^{r}

537600 + 22400e^{r} = 160000e^{r}

137600e^{r} = 537600

e^{r} = \frac{537600}{137600}

e^{r} = 3.91

Applying ln to both sides.

\ln{e^{r}} = \ln{3.91}

r = 1.36

This means that the expression for the size of the population after t years is:

P(t) = \frac{160000e^{1.36t}}{2000 + 80(e^{1.36t} - 1)}

4 0
3 years ago
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Vsevolod [243]

Answer:

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Step-by-step explanation:

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