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3 years ago

15

Geometry Question help if you can

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

3 years ago

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.

5 0

3 years ago

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:

24/6=4

1/4=6

4*6=24

24-6=18

3/4 of 24=18

3/4=75%

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