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3 years ago

9

Y=25x+35

2 answers:

kow [346]3 years ago

8 0

Answer:

The equation is in the form y=mx+b, where m=slope, b=y-intercept.

The y-intercept is 35: so the line goes through the y-axis at (0,35).

The slope is 25: so we need to count out the slope as rise over run, or 25 up and 1 to the right, starting from the y-intercept.

Slav-nsk [51]3 years ago

8 0

Y = mx + b
M is slope. b is y-intercept

The graph would start at 35 (y-int)
The slope would be counted as 25/1... aka, 25 up, 1 right.

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Read 2 more answers

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⟩

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$\theta = 108.29$

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>

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2 years ago