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sashaice [31]
3 years ago
7

Please Help I really need the answer!!!

Mathematics
1 answer:
Gre4nikov [31]3 years ago
7 0
For sketching a graph sketch on the line a trapezoid missing the bottom and the graph is continuous.
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Solve: /3=46<br><br> x/3 is a fractions
den301095 [7]

answer :138

x/3=46

x=46*3

x=138

8 0
3 years ago
Plzzzzzzzzzzzzzzzzzzzz helppppppppppppppp<br><br> 12 dived 1 1/2
balandron [24]

Answer:

8

Step-by-step explanation:

Hope this helped, Have a Wonderful Day!!

8 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
Given the following function ,find h(-4) <br><br>h(x)=6x^2+5​
Wittaler [7]

Answer:

h(-4) =101

Step-by-step explanation:

H(x) = 6x ^2 +5

We have  to find when h=4, putting it into function

h(-4) = 6(-4) ^2 +5

h(-4) = 6(16) +5

h(-4) = 96 +5

h(-4) =101

3 0
3 years ago
Mass of Weight(C)
Stolb23 [73]
I hope this helps
A.1,512
3 0
3 years ago
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