It follows from the task content that the transformation involved in the formation of the image from the pre-image is dilation by a scale factor of 1/4.

On this note, given that the coordinates of T and V from the task content are; (-4, 8) and (0,4), it follows that the coordinates of the endpoints as required are; T'(-1, 2) and V'(0, 1).

Read more on dilations;

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6 0

2 years ago

What is the standard form of the linear function that passes

babunello [35]

standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$(\stackrel{x_1}{4}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{-2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-2}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{4}}}\implies \cfrac{-3}{-2}\implies \cfrac{3}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{\cfrac{3}{2}}(x-\stackrel{x_1}{4})$

$\stackrel{\textit{multiplying both sides by }\stackrel{LCD}{2}}{2(y-1)=2\left( \cfrac{3}{2}(x-4) \right)}\implies 2y-2 = 3(x-4)\implies 2y-2=3x-12 \\\\\\ -3x+2y-2=-12\implies -3x+2y=-10\implies \stackrel{\times -1\textit{ to both sides}}{3x-2y=10}$

7 0

2 years ago

A submarine sits at –300 meters in relation to sea level. Then it descends 115 meters. What is its new position in relation to s

Dovator [93]

Answer:

-415 feet

Step-by-step explanation:

-300-115 = -415

7 0

3 years ago