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1 year ago

6

The expression cosine of pi over 2 times cosine of pi over 3 plus sine of pi over 2 times sine of pi over 3 can be rewritten as

which of the following?

1 answer:

9966 [12]1 year ago

6 0

Solution:

Given;

$\cos\frac{\pi}{2}\cos\frac{\pi}{3}+\sin\frac{\pi}{2}\sin\frac{\pi}{3}$

Using the identities;

Thus;

$\begin{gathered} \cos\frac{\pi}{2}\cos\frac{\pi}{3}+\sin\frac{\pi}{2}\sin\frac{\pi}{3}=\cos(\frac{\pi}{2}-\frac{\pi}{3}) \\ \\ \cos\frac{\pi}{2}\cos\frac{\pi}{3}+\sin\frac{\pi}{2}\sin\frac{\pi}{3}=\cos(\frac{\pi}{6}) \end{gathered}$

CORRECT OPTION: A

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a walking path across a park is represented by the equation y=-2x-7. A new path will be built perpendicular to this path.The pat

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Answer:

The equation that represents the new path is y=(1/2)x-2

Step-by-step explanation:

step 1

Find the slope of the give line

we have

y=-2x-7

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step 2

Find the slope of the perpendicular line to the given line

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal of each other

so

we have

m=-2 -----> slope of the given line

therefore

The slope of the perpendicular line is equal to

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step 3

With m=1/2 and the point (-2,-3) find the equation of the line

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

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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:

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

<u>1) Finding $(\overline{x},\overline{y})$ </u>

Average of the x coordinates:

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

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

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

$\overline{y} = 4$

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

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!

<u>3) Find the squared vertical distances between this line and the three points.</u>

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.

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

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

<u>4) Find the value of m that makes R minimum.</u>

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

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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)

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

$\dfrac{dR}{dm}=2(m-1)+0+2(2-m)(-1)$

now to find the minimum value we'll just use a condition that $\dfrac{dR}{dm}=0$

$0=2(m-1)+2(2-m)(-1)$

now solve for m:

$0=2m-2-4+2m$

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

This is the value of m for which the sum of the squared vertical distances from the points and the line is small as possible!

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3 years ago

Two numbers that differ by 46

Reil [10]

the complete question is

Find two numbers whose difference is 46 and whose product is a minimum

Let

x------->larger number

y-------> smaller number

P-------> product of the two numbers

we know that

$x-y=46\\ y=x-46$ -----> equation 1

$P=x*y$ -----> equation 2

substitute equation 1 in equation 2

$P=x*[x-46]\\ P=x^{2} -46x$

using a graph tool

see the attached figure

Find the value of x for that the product P is a minimum

the vertex is the point $(23,-529)$

that means, for $x=23$

the product is a minimum $P=-529$

find the value of y

$y=x-46\\ y=23-46\\ y=-23$

therefore

the answer is

the numbers are $23$ and $-23$

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3 years ago

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