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Natasha2012 [34]

3 years ago

12

Line $l_1$ represents the graph of $3x + 4y = -14$. Line $l_2$ passes through the point $(-5,7)$, and is perpendicular to line $

l_1$. If line $l_2$ represents the graph of $y=mx +b$, then find $m+b$.

1 answer:

nikitadnepr [17]3 years ago

5 0

Answer:

stop typing money

Step-by-step explanation:

2 timothy chpt 3 says many ppl will be lovers of money dont be one of them

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I have 2 more homework questions so please I need help.....

mafiozo [28]

hope it helps you!!!!

4 0

3 years ago

in a certain pentagon, the interior angles are a,b,c,d,e where a,b,c,d,e are integers strictly less than 180. ("Strictly less th

guajiro [1.7K]

Answer:

least to greatest: {61, 61, 61, 178, 179}

Step-by-step explanation:

If the third-largest angle is 61°, the smallest three angles cannot be larger than 183°. Since the total of all angles must be 540°, and the total of the largest two cannot be greater than 179°×2 = 358°, the sum of the smallest three must be at least 540° -358° = 182°.

So, the possible sets of angles with the smallest 3 totaling 182° or 183° are (in degrees) ...

{60, 61, 61, 179, 179} . . . . two modes

(61, 61, 61, 178, 179} . . . . . one mode -- the set you're looking for

3 0

3 years ago

Read 2 more answers

If Professor Wilson found that the test scores of his students had a variance of 4.4, what is the standard deviation? Type a num

malfutka [58]

Answer:

The standard deviation will be: 2.1

Step-by-step explanation:

We know that standard deviation is basically the square root of variance.

Using the formula to calculate the standard deviation

$s=\sqrt{s^2}$

As

The variance = s² = 4.4

so the standard deviation can be calculated as:

standard deviation $=\sqrt{s^2}$

$=\sqrt{\left(4.4\right)^}$

$=2.1$

Therefore, the standard deviation will be: 2.1

8 0

3 years ago

(12, 9, 4) What is the area of the parallelogram shown below?

zavuch27 [327]

area of the parallelogram is: b x h

12 x 4 = 48 (d)

7 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:

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

<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:

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

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)

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

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

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!

5 0

3 years ago