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kodGreya [7K]
2 years ago
11

Triangle ABC is shown with four images that are each the result of a transformation of triangle ABC. On a coordinate plane, tria

ngle G H I is reflected across the x-axis. For which triangle is one reflection necessary for the transformation to occur? Triangle G H I Triangle J K L Triangle M N P Triangle Q R S
Mathematics
2 answers:
Mice21 [21]2 years ago
5 0

Answer:

Its A

Step-by-step explanation:

Jobisdone [24]2 years ago
3 0

Answer:

a

Step-by-step explanation:

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Lines a and b are cut by transversal f. At the intersection of lines f and a, the top left angle is 96 degrees. At the intersect
Free_Kalibri [48]

Answer:

C. 22

Step-by-step explanation:

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To prove that lines a and b are parallel, equate the measures of angles 96° and (6x-36)°:

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In △ABC, m∠A=15 °, a=10 , and b=11 . Find c to the nearest tenth.
pshichka [43]

Answer:

The answer is:

\bold{c\approx 20.2\ units}

Step-by-step explanation:

Given:

In △ABC:

m∠A=15°

a=10 and

b=11

To find:

c = ?

Solution:

We can use cosine rule here to find the value of third side c.

Formula for cosine rule:

cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}

Where  

a is the side opposite to \angle A

b is the side opposite to \angle B

c is the side opposite to \angle C

Putting all the values.

cos 15^\circ = \dfrac{11^{2}+c^{2}-10^{2}}{2\times 11 \times c}\\\Rightarrow 0.96 = \dfrac{121+c^{2}-100}{22c}\\\Rightarrow 0.96 \times 22c= 121+c^{2}-100\\\Rightarrow 21.25 c= 21+c^{2}\\\Rightarrow c^{2}-21.25c+21=0\\\\\text{solving the quadratic equation:}\\\\c = \dfrac{21.25+\sqrt{21.25^2-4 \times 1 \times 21}}{2}\\c = \dfrac{21.25+\sqrt{367.56}}{2}\\c = \dfrac{21.25+19.17}{2}\\c \approx 20.2\ units

The answer is:

\bold{c\approx 20.2\ units}

4 0
2 years ago
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