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4 years ago

Write an equation that represents the line (-2,1) and (4,6)

Mathematics

2 answers:

Alexandra [31]4 years ago

6 0

Answer:

y - 6 = (5/6)(x - 4)

Step-by-step explanation:

Write an equation that represents the line through the points (-2,1) and (4,6)

As we move from (-2, 1) to (4, 6), x increases by 6 and y increases by 5.

Thus, the slope of the line connecting these two points is m = rise/run = 5/6.

Using (4, 6) as one point on the line, the equation of the line in point-slope form is

y - 6 = (5/6)(x - 4)

Basile [38]4 years ago

3 0

Answer: y - 6 = (5/6)(x - 4

Step-by-step explanation:

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Pythagorean theorem in 3D.

White raven [17]

Answer:

h = 6.2

Step-by-step explanation:

You need to find h.

The dashed triangle is a right triangle. 8 is the hypotenuse. h is a leg. The other leg is the half of the side of the base, 5.

$5^2 + h^2 = 8^2$

$h^2 + 25 = 64$

$h^2 = 64 - 25$

$h^2 = 39$

$h = \sqrt{39}$

$h = 6.244998...$

$h = 6.2$

6 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)

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

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

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

Find all the zeros of the polynomial function. x^3 + 2x^2 -5x-6 f(x) a) (-3) b) (-2, 1, 3 c) (-3, -1, 2) d) -1) e) none

Flauer [41]

Answer:1,2,3

Step-by-step explanation:

F(x)= $x^{3}$ +2 $x^2$ - $5x$ - $6$ =0

disintegrating 2 $x^2$ to $x^2$ + $x^2$

$x^{3}$ + $x^2$ + $x^2$ - $5x-6$ =0

$x^2$ $\left ( x+1\right )$ + $x^2$ -5x-6=0

$x^2$ $\left ( x+1\right )$ + $x^2$ -6x+x-6=0

$x^2$ $\left ( x+1\right )$ + $\left (x-6 \right )$ $\left ( x+1\right )$ =0

$\left ( x+1\right )$ $\left ( x^2+x-6\right )$ =0

$\left ( x+1\right )$ $\left ( x^2+3x-2x-6\right )$ =0

$\left ( x+1\right )$ $\left ( x+3\right )$ $\left ( x-2\right )$ =0

4 0

4 years ago

PLEASE HELP!!!! :D Really appreciate it if you did <3

kompoz [17]

try b

not really sure, ive never done this kind of math

5 0

3 years ago

Read 2 more answers

The table below shows the relationship between the number of calories burned and the minutes of exercise.

sergejj [24]

Answer:

Option B

Step-by-step explanation:

Since, number of calories 'C' burned is proportional to the number of minutes (m) given to the exercise,

C ∝ m

C = km

Where k is the proportionality constant.

From the table attached,

For C = 250 calories and m = 30 minutes,

250 = 30k

k = $\frac{250}{30}$

= $\frac{25}{3}$

Therefore, equation representing this situation will be,

C = $\frac{25}{3}m$

Option B will be the correct option.

5 0

3 years ago