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ra1l [238]
3 years ago
10

Evaluate k (t) = 10t-19 at the specified input. k(-7) = Explain your answer.

Mathematics
2 answers:
mr_godi [17]3 years ago
8 0
Plug -7 for x

10*-7 = -70 - 19 = -89

<span>k(-7) = -89</span>
Lera25 [3.4K]3 years ago
3 0
K(t) = 10t - 19

subtitute k = -7 to an equation:

k(-7) = 10(-7) - 19 = -70 - 19 = -89
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Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates (x_A,y_A)

Vectors AB and AB' are perpendicular, then

\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0

Vectors AC and AC' are perpendicular, then

\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0

Now, solve the system of two equations:

\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.

Subtract these two equations:

5x_A-y_A-8=0\Rightarrow y_A=5x_A-8

Substitute it into the first equation:

x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2

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Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)

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2 years ago
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