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2 years ago

What is m∠2 + m∠3? m∠2 + m∠3 =?

Mathematics

2 answers:

Serga [27]2 years ago

5 0

Let's start with m∠1. m∠1 has to be 90º because line segment AB is perpendicular with line segment AC, as it shows a right angle symbol on the angle beside m∠1. Now that we know that, it will be much easier to find m∠2 and m∠3 because m∠1, m∠2 and m∠3 are part of the triangle formed in the middle, and all angles in a triangle add up to 180º. 180-90 is 90, therefore, m∠2 +m∠3=90º so all three angles add up to 180º. You don't even have to find the specific angle measurements.

Hope this helps. I also attached a (rather poorly edited) image.

mestny [16]2 years ago

3 0

Penis. everything is just penis. penis penis penis

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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

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

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

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

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

$\theta = 108.288386087$

$\theta = 108.29$ (approximated)

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2 years ago

Which system of inequalities is shown? A. Y>x y<2 b. Y 2 D. Y>x y>2

AURORKA [14]

Answer: y > x, y > 2

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The other inequality is y > x because we shade above the dashed boundary line y = x, which is that slanted dashed line.

Combining the two regions of y > 2 and y > x leads to what is shown.

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