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1 year ago

I need some help answering this

Mathematics

1 answer:

maria [59]1 year ago

3 0

Answer:

Θ = 50°

Step-by-step explanation:

Using the cofunction identity

tan(90 - x)° = cotx°

Given

cot40°, then

tan Θ = tan(90 - 40)° = tan50°

Thus Θ = 50°

You might be interested in

50 points if you answer correctly

rewona [7]

Answer:

$f(3) = {3}^{2} + 5(3) - 9 \\ f(3) = 9 + 15 - 9 \\ f(3) = 15$

3 0

2 years ago

Read 2 more answers

Lisa spends 1/5 of an hour doing her math homework and 1/3 of an hour doing her social studies homework. What fraction of an hou

Sveta_85 [38]

To solve this, all you have to do is add 1/5 and 1/3.
Well, you might be thinking, "But the denominators are not the same! Fractions have to have like denominators in order to be added together!" And you are right. So, all we have to do, is make them both have the same denominator. To do this, we have to multiply both the numerator and denominator by the same number to find an equivalent fraction.
For 1/5, we can multiply both the numerator and denominator by 3, to get 3/15.
Likewise, we can multiply both the numerator and denominator of 1/3 by 5, to get 5/15.
Now, you can easily add 3/15 and 5/15 because all you have to do is add the numerators, because the denominators are the same!
3/15+5/15=8/15, so she spends 8/15 hour on her math and social studies homework.
Hope I helped!

3 0

3 years ago

A number written in the form a + bi is called a _____ number.

postnew [5]

The answer is complex number

a complex number is 2 parts
1. the real part
2. the imaginary part

the real part is the 'a' part of the a+bi
the imaginary part is the 'bi' part of the a+bi

answer is A
(B means exponent is no higher than 1, C means exponent is 2, D means b^2-4ac)

answer is A

5 0

3 years ago

Mr. Mole left his burrow and started digging his way down at a constant rate.

Art [367]

Mr. Mole's burrow was at an altitude of 6 meters below the ground.

Step-by-step explanation:

Step 1:

We need to determine the distance that Mr. Mole covers in a single minute.

To do that we divide the difference in values of altitude by the difference in the time periods.

For the first case, Mr. Mole had traveled -18 meters in 5 minutes.

We also have, he traveled -25.2 meters in 8 minutes.

Step 2:

The distance he covered in 1 minute $= \frac{-25.2-(-18)}{8-5} = \frac{-25.2+18}{3},$

$\frac{-25.2+18}{3} = \frac{-7.2}{3} = -2.4.$

So with every minute, Mr. Mole digs down an additional 2.4 meters below the surface.

To determine where Mr. Mole's burrow is we subtract the distance traveled in 5 minutes from -18.

The altitude of Mr. Mole's burrow $= -18 - 5(-2.4) = -18+12 = -6.$

So Mr. Mole's burrow was at an altitude of 6 meters below the ground i.e. -6 meters.

5 0

3 years ago

Read 2 more answers

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