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saveliy_v [14]
3 years ago
13

Help!!! Will give brainliest!!

Mathematics
1 answer:
Serhud [2]3 years ago
7 0

the sum of three angles of a triangle adds up to 180 degrees. In this problem, you are given two angles.

x+(2x+15)+the measure of angle Q should equal 180.

Since everything is addition, the parenthesis can be removed.

x+2x+15+Q=180. This can then be simplified to 2x+15+Q=180. Subtract 15 from both sides to get 2x+Q=165. Divide both sides by two to get the x by itself. x+Q=82.5.

Unfortunately I don't really know what to do from here, but I hope it helped at least a little.

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Let T:ℝ2→ℝ2 be the linear transformation that first rotates points clockwise through 45∘ (????/4 radians) and then reflects poin
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Answer:

T = \left[\begin{array}{ccc}-\frac{1}{\sqrt{2} } &\frac{1}{\sqrt{2} }\\\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\end{array}\right]

Step-by-step explanation:

Let General Transformation matrix be denoted as T

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General counterclockwise rotation matrix in 2-dimension is given as

                                        R(\theta)=\left[\begin{array}{ccc}cos\theta & - sin\theta\\sin\theta&cos\theta\\\end{array}\right]

For clockwise rotation we need to insert θ as negative in the above matrix. Therefore, the resulting matrix is

                                        R(-\theta)=\left[\begin{array}{ccc}cos\theta & sin\theta\\-sin\theta&cos\theta\\\end{array}\right]

as sin(-θ) = -sin (θ) and cos(-θ) = cos (θ)

For 45 degrees

sin(45)  = \frac{1}{\sqrt{2} }   and   cos(45)  = \frac{1}{\sqrt{2} }

                                       R(-45)=\left[\begin{array}{ccc}\frac{1}{\sqrt{2} }  & \frac{1}{\sqrt{2} }\\-\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\\\end{array}\right]

Step 2: Reflection through line y = x

This type of reflection maps (x,y)→(y,x)

Therefore the general matrix is

                                           R(x,y)=\left[\begin{array}{ccc}0&1\\1&0\end{array}\right]

Step 3: General Transformation Matrix

T = R(x,y) R(-θ)

                                    T=\left[\begin{array}{ccc}0&1\\1&0\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }  & \frac{1}{\sqrt{2} }\\-\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\\\end{array}\right]

                                           T = \left[\begin{array}{ccc}-\frac{1}{\sqrt{2} } &\frac{1}{\sqrt{2} }\\\frac{1}{\sqrt{2} }&\frac{1}{\sqrt{2} }\end{array}\right]

3 0
3 years ago
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