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4 years ago

Help me. I don't understand. Please help me find X and IK

Mathematics

1 answer:

sertanlavr [38]4 years ago

6 0

Answer:

$x=4\\\\y=6\\\\IK=47$

Step-by-step explanation:

Two Tangent Theorem: Tangents which meet at same point are equal in length.

Here tangents at J and H meet at I.

$Hence\ HI=IJ\\\\y^2-10=4y+2\\\\y^2-4y-12=0\\\\y^2-6y+2y-12=0\\\\y(y-6)+2(y-6)=0\\\\(y-6)(y+2)\\\\y\ can\ not\ be\ negative\ as\ in\ that\ case\ IJ\ will\ be\ negative.\\\\Hence\ y=6\\\\IJ=4y+2=4\times 6+2=26\\\\\\\\Tangents\ at\ J\ and\ L\ meet\ at\ K.\ Use\ two\ tangent\ theorem.\\\\JK=KL\\\\5x+1=6x-3\\\\6x-5x=1+3\\\\x=4\\\\JK=5x+1=5\times 4+1=21\\\\\\IK=IJ+JK=26+21\\\\IK=47$

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Find the orthogonal complement W⊥ of W and give a basis for W⊥. W = x y z : x = 1 2 t, y = − 1 2 t, z = 6t.

igor_vitrenko [27]

Answer:

W⊥= (a, b, c)

a = v - (1/2)w

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Basis for W⊥ = <(1,1,0),(-1/2,0,1)>

Step-by-step explanation:

The orthogonal complement of W is the set of vectors (a,b,c) that satisfy:

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12ta + -12tb + 6tc = 0

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a = v - (1/2)w

Therefore, the orthogonal complement W⊥ of W has the form:

W⊥ = (a,b,c) = (v - (1/2)w, v, w)

On the other hand, we can write W⊥ as:

W⊥ = (v - (1/2)w, v, w)

W⊥ = (v,v,0) + ((-1/2)w,0,w)

W⊥ = v(1,1,0) + w(-1/2,0,1)

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solniwko [45]

Answer:

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In the attached graph, we have been a little sloppy, not applying the constraints that x, y ≥ 0. For the purpose of finding the requested solution, that is of no consequence.

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