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Help I will be marking brainliest!!!

Keith_Richards [23]

<h3>Answer: A. 18*sqrt(3)</h3>

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Explanation:

We'll need the tangent rule

tan(angle) = opposite/adjacent

tan(R) = TH/HR

tan(30) = TH/54

sqrt(3)/3 = TH/54 ... use the unit circle

54*sqrt(3)/3 = TH .... multiply both sides by 54

(54/3)*sqrt(3) = TH

18*sqrt(3) = TH

TH = 18*sqrt(3) which points to choice A as the final answer

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An alternative method:

Triangle THR is a 30-60-90 triangle.

Let x be the measure of side TH. This side is opposite the smallest angle R = 30, so we consider this the short leg.

The hypotenuse is twice as long as x, so TR = 2x. This only applies to 30-60-90 triangles.

Now use the pythagorean theorem

a^2 + b^2 = c^2

(TH)^2 + (HR)^2 = (TR)^2

(x)^2 + (54)^2 = (2x)^2

x^2 + 2916 = 4x^2

2916 = 4x^2 - x^2

3x^2 = 2916

x^2 = 2916/3

x^2 = 972

x = sqrt(972)

x = sqrt(324*3)

x = sqrt(324)*sqrt(3)

x = 18*sqrt(3) which is the length of TH.

A slightly similar idea is to use the fact that if y is the long leg and x is the short leg, then y = x*sqrt(3). Plug in y = 54 and isolate x and you should get x = 18*sqrt(3). Again, this trick only works for 30-60-90 triangles.

4 0

3 years ago

If you forgot to study for a quiz with 10

sesenic [268]

I think it's a 1/16 chance

8 0

4 years ago

Warning! No identities used in the lesson may be submitted. Create your own using the columns below. See what happens when diffe

dlinn [17]

Answer:

Tthe solution is in the attached file

Step-by-step explanation:

The step by step calculation is as shown in the attachment

4 0

3 years ago

school guidelines require that there must be atleast 2 chaperones for every 25 students going on a school trip.how many chaperon

Anna007 [38]

Answer:

there must be at least 6 chaperones

Step-by-step explanation:

25 x 3= 75 so 2 chaperons x 3= 6 with a remainder of 5 students

5 0

3 years ago

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$