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3 years ago

ABCD is a rectangle. AD = 15, AC = 25, and DC = 20. Find BD.

Mathematics

1 answer:

agasfer [191]3 years ago

6 0

Answer:

Step-by-step explanation:

it is the same length as DC just going diagonal.

(i am not good at explaining in case you could not tell)

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What is the opposite of division

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Answer:

multiplication

Step-by-step explanation:

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2 years ago

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Select the correct answer.

ki77a [65]

Answer:

d 68

Step-by-step explanation:

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3 years ago

G/2h + 3f = k <br> Solve for g

Mama L [17]

Answer: if i did this correctly its {x,y} = {3,6}

Step-by-step explanation:

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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

PLEASE HELP IT WOULD BE VERY MUCH APPRECIATED! Last year, James earned £1,600 per month.

True [87]

Answer: £11.69

Explanation:

1) Work out 2% of £1600:
1600 x 2/100 =36

2) Add it on to the original number to get his new monthly pay:
1600 + 36 = £1636

3) Work out his yearly pay by:

a) finding the amount of months he worked:
46 / 4 = 11.5 months

b) Divide 11.5 into 2 bits and add them on:
1636 x 11 = 17997
1636 x 0.5 = 818

17997 + 818 = £18814

4) To get his pay per week:

18814 / 46 = £409

5) Finally, to get his pay per hour:

409 / 35 = 11.68671...
= £ 11.69

5 0

3 years ago

Read 2 more answers