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Katarina [22]
3 years ago
15

Geometry Question help if you can

Mathematics
1 answer:
leva [86]3 years ago
8 0
The answer is 18.31 because 55/4 is 13.75
Then 129 degrees minus 90 degrees is 39 degrees so
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The base of an isosceles triangle is a fourth as long as the two equal sides. write the perimeter of the triangle as a function
jeka57 [31]
Set the length of the base as x, the two equal sides are then 4x each, so the perimeter is x+4x+4x=9x
5 0
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Bob bought 3 packs of model cars he gave 4 cars to ann Bob has 11 cars left how many model cars were in each pack
Harrizon [31]
First add 11 and 4 because you need to know how many he had in total 11+4=15 then divide 15 by 3 and then you have you answer 15/3=5.
There are 5 model cars in each pack.
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If using the method of completing the square to solve the quadratic equation x² - 16x - 38 = 0, which number would have to be ad
Ray Of Light [21]

Answer:64

Step-by-Step

Image

7 0
3 years ago
PLEASE GIVE AN EXPLANATION WITH YOUR ANSWER! 6 cm are cut from a 24 cm board. What is the percent decrease in length?
Firlakuza [10]

Answer:

75%

Step-by-step explanation:

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1/4=6

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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
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