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3 years ago

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What is the domain and range of the function f(x)=−3/2(4)^(x−3)−1?

1 answer:

Mekhanik [1.2K]3 years ago

5 0

Answer:

Domain: (−∞,∞)

Range: (−∞,∞)

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What is the equation of the graph below? A graph shows a parabola that opens up with a vertex at negative three comma negative o

Tomtit [17]

Answer:

$y=(x+3)^2-1\\y=x^2+6x+8$

Step-by-step explanation:

Given:

Vertex of the parabola is, $(h,k)=(-3,-1)$

Equation of the parabola in vertex form is given as:

$y=(x-h)^2+k$

Plug in $h=-3,k=-1$ . This gives,

$y=(x-(-3))^2+(-1)\\y=(x+3)^2-1$

Therefore, the equation of the given graph is

$y=(x+3)^2-1\\y=x^2+6x+8$ .

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3 years ago

Simplify 3m/m-6 * 5m^3-30m^2/5m^2

Varvara68 [4.7K]

-30m^3-3 here is your answer

3 0

3 years ago

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What is the range and domain?

Artist 52 [7]

Answer- D-apex hope i’m right i’m not sure

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3 years ago

Find the difference (10m+3)-(4m-2)

pshichka [43]

Answer:

6m + 5

Step-by-step explanation:

3 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago