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Vanyuwa [196]
3 years ago
9

The triangles are congruent by SSS and HL. Triangles R S T and V W X are shown. Triangle R S T can be rotated about point S and

then shifted down and to the left to form triangle V W X. Which transformation(s) can be used to map △RST onto △VWX? reflection only translation only reflection, then translation rotation, then translation
Mathematics
2 answers:
NikAS [45]3 years ago
9 0

Answer:

rotation then translation

a_sh-v [17]3 years ago
7 0

Answer:

i believe its reflection then translation

Step-by-step explanation:

i think this because it you reflect it you are flipping it over and translation is just sliding it so if you reflect and translate it it will match STU

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What property is this? -9+(4+x) =(-9+4)+x
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Answer:

Associative Property

Step-by-step explanation:

The associative property lets us change the grouping, or move grouping symbols (parentheses).

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For the function y=2x^2-7 , which of these values of x corresponds to y=1
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If (5,-12) is a point on the terminal side of an angle θ, find the value of sin⁡θ, cos⁡θ, and tan⁡θ.
Bezzdna [24]

Answer:

Given point (5,-12) falls into IV quadrant

<u>The right triangle with legs of 5 and -12, and hypotenuse is:</u>

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<u>Value of sin⁡θ, cos⁡θ, and tan⁡θ:</u>

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5 0
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Can you guys help me please im so bad at math Lol so ima be posting more den one
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Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Special Triangles
Ugo [173]

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent trigonometric ratios and the exact values

cos30° = \frac{\sqrt{3} }{2} and tan30° = \frac{1}{\sqrt{3} } , then

cos30° = \frac{adjacent}{hypotenuse} = \frac{6}{m} = \frac{\sqrt{3} }{2} ( cross- multiply )

\sqrt{3} × m = 12 ( divide both sides by \sqrt{3} )

m = \frac{12}{\sqrt{3} } ← rationalise by multiplying numerator/ denominator by \sqrt{3} )

m = \frac{12}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = \frac{12\sqrt{3} }{3} = 4\sqrt{3}

-------------------------------------------------------------------------------

tan30° = \frac{opposite}{adjacent} = \frac{n}{6} = \frac{1}{\sqrt{3} } ( cross- multiply )

\sqrt{3} × n = 6 ( divide both sides by \sqrt{3} )

n = \frac{6}{\sqrt{3} } ← rationalise the denominator

n = \frac{6}{\sqrt{3} } × \frac{\sqrt{3} }{\sqrt{3} } = \frac{6\sqrt{3} }{3} = 2\sqrt{3}

-----------------------------------------------------------------------------------

7 0
3 years ago
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