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PolarNik [594]
3 years ago
8

Please help me thanks if you do

Mathematics
1 answer:
iren2701 [21]3 years ago
5 0
Problem 1 should be B, C, and D. I'm not sure about the rest... Sorry I couldn't be more of a help.
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Here are the first five terms of a sequence.
Snezhnost [94]

Step-by-step explanation:

3 to 8= 5

8 to 17= 9

17 to 30= 13

30 to 47= 17

47 to 68= 21

68 to 93= 25

93 to 122= 29

122 to 155= 33

155 to 192= 37

difference is 4

solve by

2 {n}^{2}  - n + 2 \\ 2 {(9)}^{2}  - 9 + 2 \\ 2(81) - 9 + 2 \\ 162 - 9 + 2 \\ 153 + 2 = 155

5 0
3 years ago
PLEASE HELP FAST!!! How many solutions does this system have?
Aneli [31]

Step-by-step explanation:

  • y=3/4x+(-5)
  • -4x+4(3/4x+-5)=-20

-4x+3x-20=-20

-x=-20+20

-x=0

x=0

  • x=0 in y=3/4x-5

y=3/4×{0}-5

y=-5

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}}

<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
Last year 190 student attended the spring dance and the total cost was 1250. Two years ago 175 people attended and the total cos
kondor19780726 [428]
Unit cost of a ticket = Income from ticket sales / number of tickets sold:

$1250
--------------- = $6.58 per ticket
190 tickets

Again:

$1175
--------------- = $6.71
175 tickets

While ticket prices do change (usually increase) from year to year, it's unusual to see such a situation here.

Don't have any guidelines by which to determine the "fixed cost of a ticket".

If we use the cost of a ticket of 2 years ago ($6.58/ticket), then the income from the sale of 225 tickets this year would be ($6.58/ticket)(225 tickets), or $1480.50.
6 0
4 years ago
What is the value of X?
Bad White [126]
X=12sin(60). X=10.3923...X=6rad3
5 0
3 years ago
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