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yawa3891 [41]
3 years ago
13

The triangle STU shown on the coordinate grid below is reflected once to map onto triangle S'T'U': Triangle STU drawn on a 4 qua

drant coordinate grid with vertices S is at 8, 6. T is at 2, 2. U is at 5, 1. If vertex T' is at (2, −2), what are the coordinates of vertex S'?
Mathematics
1 answer:
Andreyy893 years ago
4 0
Given that triangle <span>STU is reflected once to map onto triangle S'T'U'.
Given that triangle STU has vertices S(8, 6), T(2, 2), U(5, 1).

If vertex T' is at (2, −2), this means that triangle STU is refrected across the x-axis.

A refrection across the x-axis results in an image that has the same x-value as the pre-image but a y-value that has the opposite sign of the y-value of the pre image.

Thus, a point, say (x, y), refrected over the x-axis will result in an image with coordinate (x, -y)

Therefore, given that the coordinate of S is (8, 6), then the coordinates of vertex S'</span>  is (8, -6).
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Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?
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\mathrm{Use\:the\:rational\:root\:theorem}

a_0=15,\:\quad a_n=2

\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2

\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3

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Factor: 2x^2-11x+5

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2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)

\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:}

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x=-3,\:x=5,\:x=\frac{1}{2}

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