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How many steps does a walker average in 30 minutes?

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vampirchik [111]4 years ago

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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

3 years ago

Use the quadratic formula to solve the equation.

allochka39001 [22]

=1.3+0.1sqrt29, 1.3-0.1sqrt29

7 0

3 years ago

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the

andrew-mc [135]

Answer:

Wheres the picture?

Step-by-step explanation:

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3 years ago

Solve V=1/3bh for h. (1/3 is a fraction, idk if that matters.)

Vadim26 [7]

Answer:

3V/b = h

Step-by-step explanation:

Step 1: Write equation

V = 1/3bh

Step 2: Multiply both sides by 3

3V = bh

Step 3: Divide both sides by b

3V/b = h

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3 years ago

Find the value of x ?

Katyanochek1 [597]

Answer:

Step-by-step explanation:

x = 1.5·42 = 63

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3 years ago