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Setler79 [48]
2 years ago
8

Plot ΔABC on graph paper with points A(3,3), B(1,1), and C(6,1). Reflect ΔABC by multiplying the y-coordinates of the vertices b

y −1. Then use the function (x,y)→(x−6,y−3) to translate the resulting triangle. Name the coordinates of the vertices of the result.
Question 3 options:

A'(-3,3), B'(-1,1), C'(-1,6)


A'(-6,-3), B'(-4,-5), C'(-4,0)


A'(3,-3), B'(1,-1), C'(6,-1)


A'(-3,-6), B'(-5,-4), C'(0,-4)


Plot ΔABC on graph paper with points A(10,4), B(-1,1), and C(4,2). Reflect ΔABC by multiplying the x-coordinates of the vertices by −1. Then use the function (x,y)→(x−5,y+4) to translate the resulting triangle. Name the coordinates of the vertices of the result.

Question 4 options:

A'(-4,-10), B'(-1,1), C'(-2,-4)


A'(-8,15), B'(-5,4), C'(-6,1)


A'(-10,4), B'(1,1), C'(-4,2)


A'(-15,8), B'(-4,5), C'(-9,6)
Mathematics
1 answer:
MAXImum [283]2 years ago
6 0

Answer:

First one would be D (( A'(-3,-6), B'(-5,-4), C'(0,-4) ))

The second would be D (( A'(-15,8), B'(-4,5), C'(-9,6) )) as well

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Find the exact value of tan(x-y) if sin x=8/17 cosy = 3/5
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To solve the the question we proceed as follows:

From trigonometric laws

(cos x)^2+(sin x)^2=1

(cos x)^2+(sin x)^2=1

sin (x-y)=sin x sin y-sin y cos x

cos (x-y)=cos x cos y+sin x xin y

si  x=\frac{8}{17}

cos x=sqrt(1-(sin x)^2)=sqrt(1-64/289)=sqrt(\frac{225}{289} )=\frac{15}{17}

cos y=\frac{3}{5}

sin x= sqrt(1- (cos x)^2)= sqrt(1-\frac{9}{25} )=sqrt(\frac{16}{25} )=\frac{4}{5}

thus

tan (x-y)=[sin (x-y)]/[cos (x-y)]

=[sin x cos y-sin y cos x]/[cos x cos y+sin x sin y]

plugging in the values we obtain:

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simplifying

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