All the constants of the expression 2+6a

ASAP ILL GIVE BRAINEST!!! Find the LENGTH of Arc ABC.

saw5 [17]

Answer:

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Step-by-step explanation:

3 0

3 years ago

A breed of cattle needs at least 10 proteins and 8 fats units per day. Feed type I provides 6 protein and 2 fat units at $4 per

Vikki [24]

Let the number of bags of feed type I to be used be x and the number of bags of feed type II to be used be y, then:

We are to minimize:

C = 4x + 3y

subject to the following constraints:

$6x+2y\geq10 \\ \\ 2x+3y\geq8 \\ \\ x\geq0,\ y\geq0$

From the graph of the 4 constraints above, the corner points are (0, 5), (1, 2), (4, 0).

Testing the objective function for the minimum corner point we have:

For (0, 5):
C = 4(0) + 3(5) = $15

For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10

For (4, 0):
C = 4(4) + 3(0) = $16.

Therefore, the combination that yields the minimum cost is 1 bag of type I feed and 2 bags of type II feed.

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:

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

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)

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