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3 years ago

BRAINLIESTT ASAP! PLEASE HELP ME FAST :) thanks!

Mathematics

2 answers:

ki77a [65]3 years ago

8 0

Answer:

x = 53, y = 131

Step-by-step explanation:

Take the sum of 37 plus 90 subtract from 180 to find x

180 - (37 + 90)

180 - 127

x = 53

Since angles x and y are vertical angles, take 180 minus 53 and add 4

(180 - 53) + 4

127 + 4

y = 131

Lunna [17]3 years ago

7 0

In the figure, ∠BDC and ∠ABD are supplementary.

Supplementary angles add up to be 180°

We can write an equation symbolizing what we know:

y - 4 + x = 180

x and 37° in the figure are complementary.

A complementary angles add up to be 90°

So, we can solve for x.

x + 37 = 90

x + 37 - 37 = 90 - 37

x = 53

Substitute x with 53 and solve for y in the original equation we wrote.

y - 4 + 53 = 180

y = 180 - 53 + 4

y = 131

Therefore, the answer is [ x = 53, y = 131 ]

Best of Luck!

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nydimaria [60]

Answer:

y =7

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Step-by-step explanation:

Since this is a right triangle we can use trig functions

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tan 30 = y/ 7 sqrt(3)

7 sqrt(3) tan 30 = y

7 sqrt(3) * 1/ sqrt(3) =t

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Given equation

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Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

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