What would be the image of Point B(3, 2) after the following transformations have been applied:

Mathematics

2 answers:

Ulleksa [173]3 years ago

6 0

Answer:

(9,6)

Step-by-step explanation:

(6,9)

Because the scale factor is 3 so u r suppose to multiply by 3

FrozenT [24]3 years ago

3 0

9,6 hope this helped

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here are the dimension of the oven tray jenny bakes the cake on.600mm and 500mm.a cake tin is 25 cm in diameter .how many cake t

castortr0y [4]

First we have 2 types of units of measurement so let’s make all the same type.
600mm= 60 cm
500mm= 50 cm
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We can fit 2 rows and 2 columns of tins
Since (25+25) is less or equal to 50
And (25+25) is less or equal to 60

Answer 4 cake tins

7 0

3 years ago

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$