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3 years ago

PLZ HELP! Need this ASAP! Thx so much :)

Mathematics

1 answer:

saul85 [17]3 years ago

5 0

they are giving you the answer AB=AC, so what that means is that the angles are congruent to each other so if angle AC=15 and angle DC=5 all the angles will be the same.

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What is 60% of 64, please idk know this lol

marshall27 [118]

Its 38.4 because i am big brain

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3 years ago

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Is 7/10 less than 2/4?

lilavasa [31]

Answer:

Step-by-step explanation:

7/10 is 0.7

2/4 is 0.5

0.7>05

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3 years ago

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What is the surface area of this design ?

Fed [463]

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384 8x8=64 64x2=128 128x3=384

Step-by-step explanation:

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2 years ago

If anyone could help out with this one i would really appreciate it <br><br> f(2)=?

algol13

Answer:

f(2) = 0

Step-by-step explanation:

Function f(t) is defined in three different ways depending on the value of t.

For t = 8, function f(t) is -64/t.

For t = 10, function f(t) is 14 - t.

We are not asked about f(8) or f(10). We are asked about f(2). The third definition of function f(t) is for all values of t that are not 8 or 10. 2 is not 8 or 10, so use the third definition of function f(t) and plug in 2 for t.

f(t) = t^2 - 3t + 2 for t not equal to 8 or 10.

f(2) = 2^2 - 3(2) + 2

f(2) = 4 - 6 + 2

f(2) = 0

6 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