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jenyasd209 [6]
3 years ago
5

Jack has 10 more than twice as many baseball cards b than

Mathematics
2 answers:
tino4ka555 [31]3 years ago
8 0

Answer:

110

Step-by-step explanation:

twice as many frank has would be 100, and ten more would be 110, so jack has 110 baseball cards

Mekhanik [1.2K]3 years ago
3 0
Answer: C (110)

reasoning: 50 times 2 = 100. 100 + 10 = 110
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What is the average rate of change from x= -1 to x = 1 ?
Masteriza [31]

Answer:

  -2

Step-by-step explanation:

The average rate of change on an interval is ...

  average rate of change = (amount of change)/(width of interval)

  = (-3 -1)/(1 -(-1)) = -4/2 = -2

The average rate of change between (-1, 1) and (1, -3) is -2.

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Maggie is creating a cushion for a circular stool. The stool has a diameter of 14 inches. If she needs fabric to extend 2 inches
Marianna [84]

Answer:

396 in.^2

Step-by-step explanation:

The cushion is shaped like a cylinder with 14 inch diameter and 2 inch height.

We need to find the surface area of the cylinder.

diameter = 14 in.

radius = diameter/2 = 7 in.

height = 2 in.

A = 2(pi)r^2 + 2(pi)rh

A = 2(3.14159)(7 in.)^2 + 2(3.14159)(7 in.)(2 in.)

A = 395.84 in.^2

A = 396 in.^2

8 0
2 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
Pleas answer quickly
alexandr1967 [171]

Step-by-step explanation:

I don't understand, why there is this segment about the 0 degrees Kelvin.

any material, water or the glass the water is in, or the air, or the floor or the wall : the higher the temperature (the scale does not matter, as long both temperatures use the same), the more energy it contains (the faster the particles in that substance are moving/vibrating).

therefore a glass of water with 80 degrees will have much more energy than one with 20 degrees.

the same way an aluminum can (the actual name of that material is aluminium, but Americans seemingly cannot pronounce that) with 30 degrees has more energy than one with 20 degrees.

3 0
2 years ago
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