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Step2247 [10]
1 year ago
6

Tickets for the school football game cost $10 for students and $12 for non-students. A total of 150 ticketswere sold and $1440 w

as collected.Let x = the # of student tickets soldLet y = the # of non-student tickets sold
Mathematics
1 answer:
lyudmila [28]1 year ago
4 0

x = students = $10

y = non students = $12

total tickets = 150

Equations

x + y = 150

10x + 12y = 1440

x = 150 - y

10(150 - y) + 12y = 1440

1500 - 10y + 12y = 1440

2y = 1440 - 1500

2y = -60

y = -60/2

y = 30

x = 150 -

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Josh's annual salary is $39.000. A total of $12.330.96 will be deducted for taxes and health insurance. He will receive his payc
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Answer:

Step-by-step explanation:

without tax and insurance deduction- 3,250 (monthly)

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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

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y=m(3)-2m+4

y=m+4

vertical distance between the two points: (3,6) and (3,m+4)

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squaring:

(d_3)^2=(2-m)^2

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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}

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Serhud [2]

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b = \sqrt{\frac{C^{2} }{2(\pi )^{2} }  -  a^{2}}

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