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3 years ago

Please help finals are tomorrow

Mathematics

1 answer:

serg [7]3 years ago

3 0

The sum of internal angles in parallelogram is 360°
There are 2 angles measured x° and there are 2 angles measured 66°. So the equation will be
∠1 + ∠2 + ∠3 + ∠4 = 360°
x° + x° + 66° + 66° = 360°

Work on it
x° + x° + 66° + 66° = 360°
2x + 132 = 360
2x = 360 - 132
2x = 228
x = 228/2
x = 114

The value of x is 114

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HEL ASAP PLEASEThe dimensions of a square with side measurements of 4 inches are doubled. What happens to the area of the square

Llana [10]

Answer:

The area is multiplied by 4

Step-by-step explanation:

4x4=16 (area 1)

4(x2)x4(x2)=64 (area 2 (with the measurements doubled)

64/16=4

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3 years ago

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Write the equation of the parabola that has the vertex at point (3,−12) and passes through the point (0,6).

lana [24]

Answer:

y = 2(x-3)^2 -12

y = -4/9(x-2)^2 +7 bonus

Step-by-step explanation:

The vertex form of a parabola is

y = a(x-h)^2 + k where (h,k) is the vertex

y = a(x-3)^2 - 12

We have one point given (0,6)

6 = a (0-3) ^2 -12

6 = a (-3)^2 -12

6 = 9a-12

Add 12 to each side

6+12 = 9a

18 = 9a

Divide each side by 9

18/9 = 9a/9

a=2

y = 2(x-3)^2 -12

We follow the same steps for the bonus

y = a(x-2)^2 +7

Substitute the point into the equation

3 = a (-1-2)^2 +7

3 =a (-3)^2 +7

3 = 9a +7

subtract 7 from each side

3-7 = 9a +7-7

-4 = 9a

Divide by 9

-4/9 =a

y = -4/9(x-2)^2 +7

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2 years ago

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Which one is the unit of force A.Joules B. Newton C. Celsius D. Kelvin

Snezhnost [94]

Answer:

Newtons

Step-by-step explanation:

Joules measure electrical power and celcius and Kelvin measures temperature.

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katovenus [111]

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8 - 13 = 5

13 + 5 = 18

20 - 18 = 2

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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:

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

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)

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