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3 years ago

10

The waverly company issued 4,000 shares of common stock worth 200,000.00 total. What is the par value of each share?

1 answer:

yuradex [85]3 years ago

3 0

The value per share is 50 per share.

In order to find this, you take the total value of the stock and then divide by the number of shares.

200,000/4,000 = 50.

Therefore, it is 50 per share.

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Need to find the direction angle for each vector. #10 and #12 please!

crimeas [40]

Answer:

10) - 45 degrees

12) 0 degrees

Step-by-step explanation:

10)

Find the x and y components that define the vector that joins C with D:

x-component: -4 - (-8) = - 4 + 8 = 4

y-component: 4 - 8 = -4

use the tangent function to find the angle $\theta$ :

$tan(\theta)=\frac{y-comp}{x-comp}=\frac{-4}{4} = -1\\\theta = -45^o$

12)

Find the x and y components that define the vector that joins A with B:

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4 years ago

The question is in the picture. PLEASE HELP!!!! Thank you!!!;);)

fomenos

Answer:

19/9

Step-by-step explanation:

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3 years ago

Help. Urgent. Skenekeks

9966 [12]

Answer:

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6 0

3 years ago

Find the angle between the given vectors. Round your answer, in degrees, to two decimal places. u=⟨2,−6⟩u=⟨2,−6⟩, v=⟨4,−7⟩

NISA [10]

Answer:

$\theta = 108.29$

Step-by-step explanation:

Given

$u =$

$v =$

Required:

Calculate the angle between u and v

The angle $\theta$ is calculated as thus:

$cos\theta = \frac{u.v}{|u|.|v|}$

For a vector

$A =$

$A = a * b$

$cos\theta = \frac{u.v}{|u|.|v|}$ becomes

$cos\theta = \frac{.}{|u|.|v|}$

$cos\theta = \frac{2*6+4*-7}{|u|.|v|}$

$cos\theta = \frac{12-28}{|u|.|v|}$

$cos\theta = \frac{-16}{|u|.|v|}$

For a vector

$A =$

$|A| = \sqrt{a^2 + b^2}$

So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

$|v| = \sqrt{16+49}$

$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

$\theta = cos^{-1}(\frac{-8}{5 * 5.0990})$

$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

4 0

2 years ago