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2 years ago

The equation of line c is y–5=

Mathematics

2 answers:

Paraphin [41]2 years ago

8 0

keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the "c" equation

$y-5=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{3}{5}}(x-7)\qquad \impliedby \begin{array}{|c|ll}\cline{1-1}\textit{point-slope form}\\\cline{1-1}\\y-y_1=m(x-x_1)\\\\\cline{1-1}\end{array}$

so line "d" is a line with a slope of 3/5 and passes through (1,-1)

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{3}{5}}(x-\stackrel{x_1}{1})\implies y+1=\cfrac{3}{5}(x-1) \\\\\\ y+1=\cfrac{3}{5}x-\cfrac{3}{5}\implies y=\cfrac{3}{5}x-\cfrac{3}{5}-1\implies y=\cfrac{3}{5}x-\cfrac{8}{5}$

Fiesta28 [93]2 years ago

8 0

Answer:

y + 1 = (3/5)(x - 1)

Step-by-step explanation:

This is point-slope formula. It is super useful bc you can just fill in a point and the slope to make an equation of a line.

y - y = m(x - x)

Fill in the slope in place of the m.

Fill in the (x, y) given into the SECOND spot y and x .

Your equation had 3/5 in the m spot (that'sslope!) Swipe that and stick it in your equation.

Then they gave you a point (1, -1) that is the (x, y) you need to complete the equation.

The first y stays a y, dont fill in a number there. And the first x stays an x dont fill in a number there either.

Fill in second y, m and second x, and done! Its literally a fill in the blank question!

y + 1 = 3/5 (x - 1)

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What is the best approximation of the length of segment QS? (Note: cos 80° = 0.17)

matrenka [14]

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Step-by-step explanation:

1. You have the following information:

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EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .

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