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miss Akunina [59]
2 years ago
15

What is the slope and y-intercept of y+1=4/3x

Mathematics
2 answers:
Leona [35]2 years ago
7 0
The answer is 4/3 and the y intercept is -1
Explanation
The equation of a line is y=Mx+b so m equals the slope and you minus 1 on all side to isolate y so the y intercept or b is -1
Sergio039 [100]2 years ago
6 0

hello

so this equation is written in point slope form: y-y_{1} = m(x-x_{1})

we can solve for y to get the equation into slope intercept form: y=mx+b

y+1=4/3x

subtract 1 from both sides

y=4/3x-1

slope: 4/3

y-intercept: -1

hope this helps

have a nice day :)

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Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
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well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

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