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2 years ago

Khan academy midpoint formula question!! Please help

Mathematics

2 answers:

OlgaM077 [116]2 years ago

7 0

3,-7 is the answer for this question

LekaFEV [45]2 years ago

3 0

Answer:

The answer to your question is B = B (3, -7)

Step-by-step explanation:

Data

A = (-6, 5)

M = (-1.5, -1)

Process

1.- Write the formula of Midpoint

Xm = (x1 + x2)/2

Ym = (y1 + y2)/2

2.- Solve for x2 and y2

x2 = 2Xm - x1

y2 = 2Ym - y1

x1 = -6 y1 = 5 Xm = -1.5 Ym = -1

3.- Substitution

x2 = 2(-1.5) + 6

y2 = 2(-1) - 5

4.- Simplification

x2 = -3 + 6

y2 = -2 - 5

5.- Result

x2 = 3

y2 = -7

B (3, -7)

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2 years ago

Please help me and type out the answer

Orlov [11]

Answer:

The owner makes $2 in sales per customer

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Step-by-step explanation:

If you logically think about how the answer (now that you know it) relates to the question, it should make simple sense.

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2 years ago

Read 2 more answers

raph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the ci

andreyandreev [35.5K]

Answer:

Equation:

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

The point (0,-5), (0,7), (5,0) and (-7,0)also lie on this circle.

Step-by-step explanation:

We want to find the equation of a circle with a diamterhat hs endpoints at (-3, 4) and (5, -2).

The center of this circle is the midpoint of (-3, 4) and (5, -2).

We use the midpoint formula:

$( \frac{x_1+x_2}{2}, \frac{y_1+y_2,}{2} )$

Plug in the points to get:

$( \frac{ - 3+5}{2}, \frac{ - 2+4}{2} )$

$( \frac{ -2}{2}, \frac{ 2}{2} )$

$( - 1, 1)$

We find the radius of the circle using the center (-1,1) and the point (5,-2) on the circle using the distance formula:

$r = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

$r = \sqrt{ {(5 - - 1)}^{2} + {( - 2- - 1)}^{2} }$

$r = \sqrt{ {(6)}^{2} + {( - 1)}^{2} }$

$r = \sqrt{ 36+ 1 } = \sqrt{37}$

The equation of the circle is given by:

$(x-h)^2 + (y-k)^2 = {r}^{2}$

Where (h,k)=(-1,1) and r=√37 is the radius

We plug in the values to get:

$(x- - 1)^2 + (y-1)^2 = {( \sqrt{37}) }^{2}$

$(x + 1)^2 + (y - 1)^2 = 37$

We expand to get:

${x}^{2} + 2x + 1 + {y}^{2} - 2y + 1 = 37$

${x}^{2} + {y}^{2} + 2x - 2y +2 - 37= 0$

${x}^{2} + {y}^{2} + 2x - 2y - 35= 0$

We want to find at least four points on this circle.

We can choose any point for x and solve for y or vice-versa

When y=0,

${x}^{2} + {0}^{2} + 2x - 2(0) - 35= 0$

${x}^{2} +2x - 35= 0$

$(x - 5)(x + 7) = 0$

$x = 5 \: or \: x = - 7$

The point (5,0) and (-7,0) lies on the circle.

When x=0

${0}^{2} + {y}^{2} + 2(0) - 2y - 35= 0$

${y}^{2} - 2y - 35= 0$

$(y - 7)(y + 5) = 0$

$y = 7 \: or \: y = - 5$

The point (0,-5) and (0,7) lie on this circle.

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2 years ago

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bagirrra123 [75]

The answer is D 75% of the original volume is left.

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3 years ago

For which pair of functions is the vertex of k(x)7 units below the vertex of f(x)?

kvv77 [185]

Answer: Option C

$f(x) = x^2;\ k (x) = x ^ 2 -7$

Step-by-step explanation:

Whenever we have a main function f(x) and we want to transform the graph of f(x) by moving it vertically then we apply the transformation:

$k (x) = f (x) + b$

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units down.

If $b> 0$ then the graph of k(x) will be the graph of f(x) displaced vertically b units upwards.

In this case we have

$f (x) = x ^ 2$

We know that this function has its vertex in point (0,0).

Then, to move its vertex 7 units down we apply the transformation:

$k (x) = f (x) - 7\\\\k (x) = x ^ 2 -7$ .

Then the function k(x) that will have its vertex 7 units below f(x) is

$k (x) = x ^ 2 -7$

3 0

3 years ago