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iVinArrow [24]
3 years ago
15

Help please number 9, 10

Mathematics
1 answer:
ira [324]3 years ago
3 0

Answer:

#9, G = (-8,-10)

#10, distance = √122 = 11.0453610171

midpoint = (7/2,7/2)


Step-by-step explanation:

#9: We don't need no stinkin' diagrams!!


midpoint M=(-6,-3)

endpoint H=(-4,4)

displacement vector D from H to M:

D = (M-H) = (-6 - -4, -3 - 4) = (-2,-7)


Check: start at endpoint H, move along displacement vector D, should bring you to M.

H + D = (-4,4)+(-2,-7)

= (-4-2,4-7) = (-6,-3) = M ✔


Adding D to endpoint H gets to midpoint M. Adding 2D to H will reach the other endpoint, G.

G = H + 2D = (-4,4)+2×(-2,-7)

= (-4-2×2,4-2×7) = (-8,-10)


Check: Subtracting D from G should bring you back to M.

G-D = (-8,-10) - (-2,-7)

= (-8 - -2, -10 - -7)

= (-6,-3) = M ✔


#10 distance between P and Q, and midpoint of segment PQ:


P=(3,-2) Q=(4,9)

displacement vector D from P to Q:

D = Q - P = (4 - 3, 9 - -2) = (1,11)


so that adding D to P gives Q:

D + P = (1+3,11-2) = (4,9) ✔


Distance between two points is square root of dot product of displacement vector with itself:

d = √(D dot D) = √((1,11)dot(1,11))

= √(1×1+11×11) = √122 = 11.0453610171


Midpoint M is P + D/2, start at P and move half way to Q,

M = (3,-2)+(1,11)/2

= (3+1/2,-2+11/2)

= (7/2,7/2)

Start at M, move D/2 brings you to Q,

M + D/2 = (7/2+7/2) + (1/2,11/2)

= (8/2,18/2) = (4,9) = Q ✔


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On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)
xz_007 [3.2K]

Answer:

P = 10 + 2\sqrt{53} units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3),  X (2, 3),

Y (4, -4),  Z (-3, -2)

Required

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First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

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For XY:

(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)

XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}

XY = \sqrt{-2^2 + (3 +4)^2}

XY = \sqrt{-2^2 + 7^2}

XY = \sqrt{4 + 49}

XY = \sqrt{53}

For YZ:

(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)

YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}

YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}

YZ = \sqrt{7^2 + (-2)^2}

YZ = \sqrt{49 + 4}

YZ = \sqrt{53}

For ZW:

(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)

ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}

ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}

ZW = \sqrt{0^2 + (-5)^2}

ZW = \sqrt{0 + 25}

ZW = \sqrt{25}

ZW = 5

The Perimeter (P) is as follows:

P = WX + XY + YZ + ZW

P = 5 + \sqrt{53} + \sqrt{53} + 5

P = 5 + 5 + \sqrt{53} + \sqrt{53}

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6 0
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Answer:

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