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Natasha_Volkova [10]
3 years ago
12

Rectangle ABCD is dilated by a scale factor of 1/2 with a center of dilation at the origin. What are the coordinates of the imag

e of point B'?
A. (–2, 3)
B. (3, –2)
C. (–8, 12)
D. (12, –8)

Mathematics
2 answers:
Alex Ar [27]3 years ago
6 0
A. (-2,3) because Rectangle ABCD is dilated by a scale factor of 1/2.
docker41 [41]3 years ago
6 0

Answer:

The correct option is A.

Step-by-step explanation:

If a figure dilated by a scale factor of k with a center of dilation at the origin, then

(x,y)\rightarrow (kx,ky)

It is given that rectangle ABCD is dilated by a scale factor of 1/2 with a center of dilation at the origin, so

(x,y)\rightarrow (\frac{1}{2}x,\frac{1}{2}y)

The coordinates of B are (-4,6). So, the coordinates of B' are

B(-4,6)\rightarrow B'(\frac{1}{2}(-4),\frac{1}{2}(6))

B(-4,6)\rightarrow B'(-2,3)

The coordinates of B' are (-2,3), therefore option A is correct.

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\frac{12}{15}\\\\divide\ a\ numerator\ and\ a\ denominator\ by\ 3\\\\12:3=4;\ 15:3=5\\\\therefore\ \frac{12}{15}=\frac{4}{5}\\\\multiply\ a\ numerator\ and\ a\ denominator\ by\ 2\\\\4\cdot\ 2=8;\ 5\cdot2=10\\\\therefore\ \frac{12}{15}=\frac{4}{5}=\frac{8}{10}=\boxed{0.8}\Rightarrow0.8\cdot100\%=\boxed{80\%}



\frac{68}{80}\\\\divide\ a\ numerator\ and\ a\ denominator\ by\ 4\\\\68:4=17;\ 80:4=20\\\\therefore\ \frac{68}{80}=\frac{17}{20}\\\\multiply\ a\ numerator\ and\ a\ denominator\ by\ 5\\\\17\cdot\ 5=85;\ 20\cdot5=100\\\\therefore\ \frac{68}{80}=\frac{17}{20}=\frac{85}{100}=\boxed{0.85}\Rightarrow0.85\cdot100\%=\boxed{85\%}
7 0
3 years ago
Read 2 more answers
If p=(-3,-2) and q=(1,6) are the endpoints of the diameter of a circle find the equation of the circle
stira [4]

Answer:

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<u>Step-by-step explanation:</u>

Given points end  Points are p(-3,-2) and q( 1,6)

<em>The distance of two points formula</em>

P Q = \sqrt{x_{2} - x_{1})^{2} + ((y_{2} -y_{1})^{2}   }

P Q = \sqrt{1 - (-3)^{2} + ((6 -(-2))^{2}   }

P Q = \sqrt{16+64} = \sqrt{80}

<em>The diameter 'd' = 2 r</em>

                       2 r = √80

                            = \sqrt{16 X 5}

                           = 4 \sqrt{5}

                     <em> r =  2√5</em>

<em>Mid-point of two end points </em>

<em>                        </em>(\frac{x_{1} + x_{2} }{2} , \frac{y_{1} +y_{2} }{2} ) = (\frac{-3+1}{2} ,\frac{-2 +6}{2} )<em></em>

<em>                                               = (-1 ,2)</em>

<em>Mid-point of two end points = center of the circle</em>

<em>                                     (h,k) = (-1 , 2)</em>

                 

The equation of the circle

           (x -h )² +(y-k)² = r²

<em>          (x -(-1) )² +(y-(2))² = (2(√5))²</em>

<em>         x² + 2 x + 1 + y² - 4 y + 4 = 20</em>

<em>          x² + 2 x + y² - 4 y  = 20 -5</em>

<em>           x² + 2 x + y² - 4 y  = 15</em>

<u><em>Final answer</em></u><em>:-</em>

<em>The equation of the circle   (x +1) )² +(y-(2))² = (2(√5))²</em>

<em>   or </em>

<em>The equation of the circle    x² + 2 x + y² - 4 y  = 15</em>

<em>                  </em>

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Isolate the variable by dividing each side by factors that don't contain the variable.
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6 < x+5\leq11\qquad|\text{subtract 5 from every sides}\\\\1 < x\leq6\to x\in(1,\ 6]

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