If m∠4=6y and m∠5=13y-10, find m<6. WILL GIVE YOU Brainliest

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

3 0

Answer:

$m\angle 1 = 115\degree \\\\m\angle 2= 130\degree \\\\m\angle 6= 60\degree \\\\$

Step-by-step explanation:

$m\angle 1 = m \angle 5\\(corresponding \: \angle s) \\\\\because m\angle 5 = 115\degree \\\therefore m\angle 1 = 115\degree \\\\m\angle 2 = m \angle 7\\(exterior \: alternate \: \angle s) \\\\\because m\angle 7 = 130\degree \\\therefore m\angle 2= 130\degree \\\\m\angle 4 + m\angle 5= 180\degree\\\\6y\degree + (13y - 10)\degree = 180\degree\\\\ (19y - 10)\degree = 180\degree\\\\ 19y - 10 = 180\\\\19y = 180+ 10\\\\19y = 190\\\\y = \frac{190}{19}\\\\y = 10\\\\m \angle 4 = 6y\degree \\\\m \angle 4 = 6\times 10\degree \\\\m \angle 4 = 60\degree \\\\\because m\angle 6 = m \angle 4\\( alternate \: \angle s) \\\\\therefore m\angle 6= 60\degree \\\\$

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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, $9+\sqrt{185}$ .