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3 years ago

Complete the proof.....

Mathematics

1 answer:

Zigmanuir [339]3 years ago

3 0

Answer:

Step-by-step explanation:

From the given attachment to the question, the following can be deduced.

1. M is the midpoint of GH, given that M is the definition of midpoint.

2. GM ≅ MH, given that x = 2. So that,

GM = 5x = 5(2)

= 10

MH = 3x + 4 = 3(2) + 4

= 10

Thus, GM ≅ MH

3. GM = MH is the definition of congruence. This implies that they are similar.

4. 5x = 3x + 4, by substitution property of equality.

Since, x = 2 then;

5(2) = 3(2) + 4

10 - 10

5. 2x = 4, given that x = 2.

6. GH = GM + MH, by division property of equality.

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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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method:

take the first one - four eighths, which is 4/8

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Read 2 more answers

Please help ASAP don’t mind the scribbles

hjlf

U = 1

L = 0

R = -2

I = 24

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H = -3

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S = -439

O = 11

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T = -41

D = 19

C = 206

<em>Hope this helped ! :)</em>

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guapka [62]

Y = 3x - 1

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y = mx + b

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y = 3x + b

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