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2 years ago

The coordinates of the three vertices of the triangle are A (x1, y1), B (x2, y2), C (x3, y3).

Mathematics

1 answer:

kaheart [24]2 years ago

6 0

Answer:

Step-by-step explanation:

Using the section formula, if a point (x,y) divides the line joining the points (x

) and (x

) in the ratio m:n, then

(x,y)=(

m+n

+nx

m+n

+ny

)

The vertices of the triangle are given to be (x

),(x

) and (x

). Let these vertices be A,B and C respectively.

Then the coordinates of the point P that divides AB in l:k will be

(

l+k

+kx

l+k

+ky

)

The coordinates of point which divides PC in m:k+l will be

⎩

⎪

⎨

⎪

⎧

m+k+l

+(k+l)

(l+k)

+kx

m+k+l

+(k+l)

(l+k)

+ky

⎭

⎪

⎬

⎪

⎫

⇒(

m+k+l

+lx

+mx

m+k+l

+ly

+my

)

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Degger [83]

Answer:

Step-by-step explanation:

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2 years ago

Please help! I get how to do the first part I just don't get how to do the second part.

xz_007 [3.2K]

9514 1404 393

Answer:

WX = 33

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Step-by-step explanation:

The hash marks tell you WX is a midline, so has the measure of the average of the two bases.

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The hash marks also tell you ...

PW = WS

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and also

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Substituting that into the second equation gives ...

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2 years ago

Simplify these expressions:<br>5m^2 x 2m<br>60cf/3f<br>15c-5f/2

Mars2501 [29]

5m^2 x 2m = 10m^3
60cf/3f = 20cf^2
15c-5f/2 = Cannot be simplified further

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2 years ago

To the nearest tenth, find the perimeter of ABC with vertices A (-2,-2) B (0,5) and C (3,1)

Pavlova-9 [17]

the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}$

$\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill$

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2 years ago

Intr un album numarul de fotografii s a micsorat cu 50%. Cu cat trebuie sa se mareasca pentru a reveni la numarul initial? Raspu

hodyreva [135]

Răspuns:

100%

Explicație pas cu pas:

Permiteți numărul inițial de fotografii = x

x - 50% din x = x - 0,5x = 0,5x

0,5x + câte% din 0,5x = x

x - 0,5x = y% de 0,5x

0,5x / 0,5x = 1

Prin urmare,

100% = 1

Prin urmare, x trebuie mărit cu 100% pentru a reveni la numărul original.

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2 years ago