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Find the coordinates of the midpoint of the segment whose endpoints are H(8, 13) and K(10, 9)

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Answer:

The coordinates of the mid point  the line segment whose endpoints are H(8, 13) and K(10, 9) is (9, 11)                                          

Step-by-step explanation:

Given: A line segment whose endpoints are H(8, 13) and K(10, 9)  

We have to find the mid point of the line segment whose endpoints are H(8, 13) and K(10, 9).

Consider the given end points H(8, 13) and K(10, 9).

Using Formula for mid point,

M=left(frac{x_2+x_1}{2},::frac{y_2+y_1}{2}right)

We have,

left(x_1,:y_1right)=left(8,:13right),:left(x_2,:y_2right)=left(10,:9right)

Substitute, we have,

M=left(frac{10+8}{2},:frac{9+13}{2}right)

Simplify, we have,

M=left(9,:11right)

Thus, The coordinates of the mid point  the line segment whose endpoints are H(8, 13) and K(10, 9) is (9, 11)

                                         

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Identify both the x- and y-intercepts of the linear equation graphed on the coordinate plane above. Use the intercepts to write an equation of the line in point-slope form, slope-intercept form, and general form of a linear equation.Complete your work in the space provided or upload a file that can display math symbols if your work requires it. In your work, be sure to include the coordinates for both intercepts and the equations of the line in all three formats.

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The x int (where the line crosses the x axis) is (-2,0)
the y int (where the line crosses the y axis) is (0,-2)

(-2,0)(0,-2)
slope = (-2 – 0) / (0 – (-2) = -2/2 = -1

y = mx + b
slope(m) = -1
use either of ur points (-2,0)…x = -2 and y = 0
sub and find b
0 = -1(-2) + b
0 = 2 + b
-2 = b

so ur equation in slope intercept form is : y = -x – 2

y = -x – 2
x + y = -2
x + y + 2 = 0 <== general form

there can be 2 answer for point slope form…
y – y1 = m(x – x1)
slope(m) = -1
(-2,0)…x1 = -2 and y1 = 0
sub
y – 0 = – (x – (-2)
y – 0 = -(x + 2) <=== point slope form

y – y1 = m(x – x1)
slope(m) = -1
(0,-2)…x1 = 0 and y1 = -2
sub
y – (-2) = – (x – 0)
y + 2 = – (x – 0) <== point slope form

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What is the most specific name that can be given to a figure with the following coordinates? (–10, 8), (–7, 13), (3, 7), and (0, 2)

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What is the most specific name that can be given to a figure with the following coordinates? (–10, 8), (–7, 13), (3, 7), and (0, 2)

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set of coordinates, when paired with (-3, -2) and (-5, -2), result in a square

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set of coordinates, when paired with (-3, -2) and (-5, -2), result in a square

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JKLM is translated according to the rule (x,y) (x+5,y-4) what are the coordinates of .0

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JKLM is translated according to the rule (x,y) (x+5,y-4) what are the coordinates of .0

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Express the polar coordinates (-3, pi/2) as Cartesian coordinates. Choices: (-3,0) (-3,3) (0,-3) (3,-3)

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Express the polar coordinates (-3, pi/2) as Cartesian coordinates. Choices: (-3,0) (-3,3) (0,-3) (3,-3)

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________ software helps run the computer and coordinates instructions between other software and the hardware devices.

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Cloud computing infrastructure model provides access to shared computing resources over a network, often the internet. The cloud computing infrastructure is a virtual infrastructure composed of hardware and software components needed to support the computing requirements of a cloud computing model.And the cloud computing model is the delivery of computing services (servers, storage, databases, networking, software, analytics and more) over the Internet . 

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WILL GIVE BRAINLIEST!A rhombus has coordinates A(-6, 3), B(-4, 4), C(-2, 3), and D(-4, 2). What are the coordinates of rhombus A′B′C′D′ after a 90° counterclockwise rotation about the origin followed by a translation 3 units to the left and 2 units down? A′(-6, -8), B′(-7, -6), C′(-6, -4), D′(-5, -6) A′(6, -8), B′(7, -6), C′(6, -4), D′(5, -6) A′(-6, 8), B′(-7, 6), C′(-6, 4), D′(-5, 6) A′(6, 8), B′(7, 6), C′(6, 4), D′(5, 6)

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Answer:

C. Yes, because the population values appear to be normally distributed.

Step-by-step explanation:

Given is a graph which shows the distribution of values of a population

The graph has the following characteristics

i) Bell shaped

ii) symmerical about mid vertical line

iii) Unimodal with mode = median =mean

iv) As x deviates more from the mean probability is decreasing and also curve approaches asymptotically the x axis

Hence we find that the curve is a distribution of normal

Option C is right

C. Yes, because the population values appear to be normally distributed.

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Parallelogram abcd has vertex coordinates A (-4, 4), B (-1, 4), C (-2, 2), D (-5,2). it is reflected across the y- axis. what are the coordinates of A? A. (-4, 4), B. (4, -4), C. (-4, −4), D. (-5, 2) plz help anyone

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Parallelogram abcd has vertex coordinates A (-4, 4), B (-1, 4), C (-2, 2), D (-5,2). it is reflected across the y- axis. what are the coordinates of A? A. (-4, 4), B. (4, -4), C. (-4, −4), D. (-5, 2) plz help anyone

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What are the coordinates of the image of vertex R after a reflection across the y-axis?

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What are the coordinates of the image of vertex R after a reflection across the y-axis?

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In polar coordinates the positive x axis is called?

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Answer:

3 positive real root and 1 negative real root and no complex root.

Step-by-step explanation:

Here, the given function,

f(x) = -3x^4 + 5x^3 - x^2 + 8x + 4

Since, the coefficient of variables are,

-3,   5,    -1,    8,    4,

The sign of variables goes from Negative(-3) to positive (5) , positive(5) to negative(-1) and negative (-1) to positive (8),

So, the total changes in sign = 3,

By the Descartes’s rule of sign,

Hence, the number of real positive roots = 3,

Also,

f(-x) = -3x^4 - 5x^3 - x^2 - 8x + 4

Since, the coefficient of variables are,

-3,   -5,   -1,    -8,   4

The sign of varibles goes to negative (-8) to positive (4),

So, the total changes in sign = 1,

Hence, the number of real negative roots = 1,

Now, the degree of the function f(x) is 4,

Thus, the highest number of roots of f(x) is 4,

So, it does not have any complex root.

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Find the midpoint between two points on a coordinate plane whose coordinates are (5, 7) and (-3, 1). (2, 8) (1, 4) (8, 8) (4, 4)

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Find the midpoint between two points on a coordinate plane whose coordinates are (5, 7) and (-3, 1). (2, 8) (1, 4) (8, 8) (4, 4)

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The midpoint of MN is point P at (–4, 6). If point M is at (8, –2), what are the coordinates of point N?

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Answer:  (-16, 14)

Step-by-step explanation:

We know that the coordinates of mid point of a line segment having end points (a,b) and (m,n) is given by :-

x=dfrac{a+m}{2}  ,  y=dfrac{b+n}{2}

Given : The midpoint of MN is point P at (-4, 6). If point M is at (8, -2).

Let the coordinates of N be (a,b).

Then by using the above formula, we have

-4=dfrac{8+a}{2}  ,  6=dfrac{-2+b}{2}

Rightarrow -8=8+a  ,  12=-2+b   [Multiplying 2 on both sides]

Rightarrow a=-8-8=-16  ,  b=12+2=14

Hence, the coordinates of point N = (-16, 14)

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Find the coordinates of the circumcenter for ∆DEF with coordinates D(1,1) E (7,1) and F(1,5). Show your work

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Answer:  The correct option is (C). 10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared

Step-by-step explanation:  Given that the segment AB has point A located at (8, 9). The distance from A to B is 10 units.

We are to select the correct option that could be used to calculate the coordinates for point B.

Let, (x, y) be the co-ordinates of point B.

According to distance formula, the distance between two points (a, b) and (c, d) is given by

d=sqrt{(c-a)^2+(d-b)^2}.

Therefore, the distance between the points A(8, 9) and B(x, y) is given by

d=sqrt{(x-8)^2+(y-9)^2}.

Since, distance between A and B is 10 units, so

d = 10.

Therefore,

10=sqrt{(x-8)^2+(y-9)^2}.

Thus, the correct statement is

10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared.

Option (C) is correct.

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Find the coordinates of point l that lies along the directed line segment from N(14,4) to M(2,8) and partitions the segment in the ratio of 1:3

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1. Drop the perpendicular from M to the x -axis, and the perpendicular from N to the y-axis.

Let these perpendiculars intersection be point A, with coordinates (2, 4).

[clearly A will have x coordinate equal to the x coordinate of M, and y coordinate equal to the y coordinate of N]

2. Let P be the point on NM such that |NP|/|PM|=1/3

3. Drop the perpendiculars from point P to AN and AM. Let the foot of these perpendiculars be points P’ and P” respectively.

4. By Thales theorem: |NP’|/|P’A|=1/3 and |AP”|/|P”M|=1/3

so let |NP’|=b, |P’A|=3b, |AP”|=c, |P”M|=3c, as shown in the figure

5. |AN|=14-2=12. 
    |AN|=4b so b=12/4=3

the x coordinate of P’=2+3b=2+9=11
the x coordinate of P= the x coordinate of P’=11

similarly |AM|=8-4=4
              |AM|=4c so c=1

the y coordinate of P” is 4+c=4+1=5

the y coordinate of P = the y coordinate of P”=5

Answer: (11, 5)

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True or False – One advantage of polar equations is that you can use a polar function to describe a graph whose equation in rectangular coordinates would not be considered a function.

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Answer:  The correct option is (C). 10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared

Step-by-step explanation:  Given that the segment AB has point A located at (8, 9). The distance from A to B is 10 units.

We are to select the correct option that could be used to calculate the coordinates for point B.

Let, (x, y) be the co-ordinates of point B.

According to distance formula, the distance between two points (a, b) and (c, d) is given by

d=sqrt{(c-a)^2+(d-b)^2}.

Therefore, the distance between the points A(8, 9) and B(x, y) is given by

d=sqrt{(x-8)^2+(y-9)^2}.

Since, distance between A and B is 10 units, so

d = 10.

Therefore,

10=sqrt{(x-8)^2+(y-9)^2}.

Thus, the correct statement is

10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared.

Option (C) is correct.

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Find the coordinates of point Q that lies along the directed line segment from R(-2, 4) to S(18, -6) and partitions the segment in the ratio of 3:7. Please help!!

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Find the coordinates of point Q that lies along the directed line segment from R(-2, 4) to S(18, -6) and partitions the segment in the ratio of 3:7. Please help!!

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Find the coordinates of point Q that lies along the directed line segment from R(-2, 4) to S(18, -6) and partitions the segment in the ratio of 3:7.

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Find the distance between them
distance between (x1,y1) and (x2,y2) is
D=sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

so distance between (-2,4) and (18,-6) is
D=sqrt{(18-(-2))^2+(-6-4)^2}
D=sqrt{(18+2)^2+(-10)^2}
D=sqrt{(20)^2+100}
D=sqrt{400+100}
D=sqrt{500}
D=10√5

so in ratio 3:7
3+7=10
3/10 of 10√5 is 3√5

I’m thinkin you want R then Q then S such that RQ:QS=3:7
so
distance from R to Q is 3√5
Q is (x,y)
R is (-2,4)
D=3√5

3sqrt{5}=sqrt{(-2-x)^2+(4-y)^2}
3sqrt{5}=sqrt{x^2+4x+4+y^2-8y+16}
square both sides
45=x^2+4x+4+y^2-8y+16
45=x^2+y^2+4x-8y+20

now, ithas to be on the line that R and S are on
do some simple math
slope is rise/run=-10/20=-1/2
y=-1/2x+b
4=-1/2(-2)+b
4=1+b
3=b
y=-1/2x+b

sub that for y in our other equation (45=x^2+y^2+4x-8y+20)

45=x^2+(frac{-1}{2}x+3)^2+4x-8(frac{-1}{2}x+3)+20
I’m too lazy to show you expanstion and whatnot so I’ll give you the solution
we get (after some manipulation)
0=x²+4x-32
what  2numbers multiply to get -32 and add to get 4?
-4 and 8
0=(x-4)(x+8)
set to zero

0=x-4
4=x

0=x+8
-8=x
but wait, -8 is not between -2 and 18 so it can’t be

so x=4

remember, y=-1/2x+3
sub that to get y=-1/2(4)+3=-2+3=1

the point is (4,1)
apologies for mishap

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The midpoint of a segment is (4,3) and one endpoint is (10,8) Find the coordinates of the other endpoint.

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Answer:  The correct option is (C). 10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared

Step-by-step explanation:  Given that the segment AB has point A located at (8, 9). The distance from A to B is 10 units.

We are to select the correct option that could be used to calculate the coordinates for point B.

Let, (x, y) be the co-ordinates of point B.

According to distance formula, the distance between two points (a, b) and (c, d) is given by

d=sqrt{(c-a)^2+(d-b)^2}.

Therefore, the distance between the points A(8, 9) and B(x, y) is given by

d=sqrt{(x-8)^2+(y-9)^2}.

Since, distance between A and B is 10 units, so

d = 10.

Therefore,

10=sqrt{(x-8)^2+(y-9)^2}.

Thus, the correct statement is

10 = square root of the quantity of x minus 8 all squared plus y minus 9 all squared.

Option (C) is correct.

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What are the coordinates of a point on the unit circle if the angle formed by the positive x- axis and the radius is 45°??

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What are the coordinates of a point on the unit circle if the angle formed by the positive x- axis and the radius is 45°??

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The slope is -8/7 and the coordinates are (4,4).Whats the answer?

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The basis to respond this question are:

1) Perpedicular lines form a 90° angle between them.

2) The product of the slopes of two any perpendicular lines is – 1.

So, from that basic knowledge you can analyze each option:

a.Lines s and t have slopes that are opposite reciprocals.

TRUE. Tha comes the number 2 basic condition for the perpendicular lines.

slope_1 * slope_2 = – 1 => slope_1 = – 1 / slope_2, which is what opposite reciprocals means.

b.Lines s and t have the same slope.

FALSE. We have already stated the the slopes are opposite reciprocals.

c.The product of the slopes of s and t is equal to -1

TRUE: that is one of the basic statements that you need to know and handle.

d.The lines have the same steepness.

FALSE: the slope is a measure of steepness, so they have different steepness.

e.The lines have different y intercepts.

FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.

f.The lines never intersect.

FALSE: perpendicular lines always intersept (in a 90° angle).

g.The intersection of s and t forms right angle.

TRUE: right angle = 90°.

h.If the slope of s is 6, the slope of t is -6

FALSE. – 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is – 1/6.

So, the right choices are a, c and g.

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The center of a circle on a cordinate plane is located at the point (2,-1). one endpoint of the diameter is located (0,-2). what are the coordinates of the other endpoint of the diamtere of the circle

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Answer:

                The point that lie on the line is:

                        B.  (1,1)

Step-by-step explanation:

We are given that a line passes through the point (0,-1) and has a slope of 2.

We know that the equation of a line passing through (a,b) and having slope m is given by:

           y-b=m(x-a)

Here we have:  (a,b)=(0,-1) and m=2

This means that the equation of line is:

y-(-1)=2(x-0)\\y+1=2x\\y=2x-1

Now we will check which option is true.

A)

                (2,1)

when x=2

we have:

y=2times 2-1\\\y=4-1\\\y=3neq 1

Hence, option: A is incorrect.

B)

                     (1,1)

when x=1

we have:

y=2times 1-1\\\y=2-1\\\y=1

                Hence, option: B is correct.

C)

                      (0,1)

when x=0

we have:

y=2times 0-1\\\y=0-1\\\y=-1neq 1

Hence, option: C is incorrect.

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Three trigonometric functions for a given angle are shown below. Cosecant theta equals 13/12 secant Theta equals -13/5, cotangent equals -5/12. Where are the coordinates of point (x,y) On the terminal ray of angle theta assuming that the values above not simplified ?

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Answer:

False

Step-by-step explanation:

Suposse that we are given a function f(x) and a constant value h.

1. Case:

If we take the function g(x)=f(x)+h, then the graph of the function g(x) will be the graph of the funcion f(x) moved up or down.

2.Case:

If we take the function g(x)=hf(x), then the graph of the function g(x) will be the graph of the function f(x) just taller or shorter.

3.Case:

If we take the function g(x)=f(x-h), then the graph of the function g(x) will be the graph of the fuction f(x) moved horizontally.

4. Case:

If we take the function g(x)=f(hx), then the graph of the function g(x) will be tha graph of the function f(x) wither or thiner.

For example:

If we take f(x)=sin(x) and h=2. Then, if we take g(x)=sin(2x) then f(0)=g(0)=0, which means that the graph of the functiction is not moved up or down. However, f(π/2)=sin(π/2)=1 and g(π/2)=sin(π)=0 which gives us a hint that the graph of the function became thiner.

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Which of the following coordinates of point C would make triangle ABC similar to triangle LMP ? Look at pic attached :) thank you .

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Answer:

                The point that lie on the line is:

                        B.  (1,1)

Step-by-step explanation:

We are given that a line passes through the point (0,-1) and has a slope of 2.

We know that the equation of a line passing through (a,b) and having slope m is given by:

           y-b=m(x-a)

Here we have:  (a,b)=(0,-1) and m=2

This means that the equation of line is:

y-(-1)=2(x-0)\\y+1=2x\\y=2x-1

Now we will check which option is true.

A)

                (2,1)

when x=2

we have:

y=2times 2-1\\\y=4-1\\\y=3neq 1

Hence, option: A is incorrect.

B)

                     (1,1)

when x=1

we have:

y=2times 1-1\\\y=2-1\\\y=1

                Hence, option: B is correct.

C)

                      (0,1)

when x=0

we have:

y=2times 0-1\\\y=0-1\\\y=-1neq 1

Hence, option: C is incorrect.

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Triangle PQR has vertices at the following coordinates: P(0, 1), Q(3, 2), and R(5, -4). Determine whether or not triangle PQR is a right triangle. Show all calculations for full credit.

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Triangle PQR has vertices at the following coordinates: P(0, 1), Q(3, 2), and R(5, -4). Determine whether or not triangle PQR is a right triangle. Show all calculations for full credit.

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