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3 years ago

7

Consider the trapezoid shown below. 6 in. 4 in. 6 in. 6 in. (not drawn to scale)

1 answer:

AnnyKZ [126]3 years ago

7 0

Answer:

36

Step-by-step explanation:

1/2bh

1/2 (6*4) =12

Considering there are three triangles we will multiply by 3

12*3=36

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What is 4 percent of 1425

hammer [34]

4% outta 1425 is 57| The amount is 57.

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4 years ago

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Let 2004-02-01-02-00_files/i0200000.jpg. If 2004-02-01-02-00_files/i0200001.jpg is a complement of 2004-02-01-02-00_files/i02000

Reika [66]

Answer: The answer is (C) 180°.

Step-by-step explanation: Given that m∠A = 40°. ∠B is a complement of ∠A and ∠C is a supplement of ∠B. We are to find m∠B + m∠C.

Since ∠B is a complement of ∠A, so we have

$m\angle A+m\angle B=90^\circ\\\\\Rightarrow m\angle B=90^\circ-m\angle A=90^\circ-40^\circ\\\\\Rightarrow m\angle B=50^\circ.$

Now, since ∠C is a supplement of ∠B, so we have

$m\angle B+m\angle C=180^\circ\\\\\Rightarrow m\angle C=180^\circ-m\angle B=180^\circ-50^\circ\\\\\Rightarrow m\angle C=130^\circ.$

Therefore,

$m\angle B+m\angle C=50^\circ+130^\circ=180^\circ.$

Thus, (C) is the correct answer.

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3 years ago

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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:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

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3 years ago

Solve 3[-x + (2 x + 1)] = X – 1.<br> Х=

Strike441 [17]

Answer:

x=-2

Step-by-step explanation:

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3 years ago

A jet fighter is capable of reaching speeds up to 1.5 x 10^3 miles per hour. A typical passenger plane can reach speeds of appro

Volgvan

Answer:

3 times faster

Step-by-step explanation:

$\frac{1.5(10)^{3} }{5.0(10)^{2}} = \frac{1.5(10)}{5} = \frac{15}{5} = 3$ times faster

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3 years ago