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2 years ago

Draw the triangle with the given vertices. Multiply each coordinate of the vertices by 3, and then draw the new triangle.

Mathematics

2 answers:

shtirl [24]2 years ago

8 0

(-9,18) (3,6) (3,-12) should be the new coordinates

geniusboy [140]2 years ago

3 0

Answer:

(-9,6), (3,6), (3,-12) The dilation is an enlargement.

Step-by-step explanation:

-3x3=-9

2x3=6

1x3=3

2x3=6

1x3=3

-4x3=-12

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2 years ago

Find a quadratic function in standard form for each set of points

otez555 [7]

For 1 the equation of the parabola is y = x^2 - 8x + 3
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2 years ago

a rectangle has a width of 4 cm and a length of 8cm is scaled by a factor of 6.what are the side lengths of the scaled copy

vichka [17]

Answer:

$\huge\boxed{\text{Width: } 24 \ \ \ \text{Length: } 32}$

Step-by-step explanation:

When we scale an object by a scale factor, the side lengths become the scale factor times bigger than the previous sides. Therefore, with a scale factor of 6, the side lengths would be multiplied by 6.

$4 \cdot 6 = 24\\\\4 \cdot 8 = 32$

Hope this helped!

6 0

3 years ago

Read 2 more answers

Use an area model to solve. 1 3/4 X 2 1/2

Sauron [17]

I can give you the answer right of the top of my head here it is 1 and 3/4 x 2 and 1/2 is 4 and 3/8

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3 years ago

Write the equation of the line that is the perpendicular bisector of the segment with endpoints (4, 1) and (2, -5)

Sedbober [7]

Answer: $\bold{y=-\dfrac{1}{3}x-1}$

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

(4, 1) & (2, -5)

First, find the slope (m) and then the perpendicular (opposite reciprocal) slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\\m=\dfrac{-5-1}{2-4} = \dfrac{-6}{-2}=3\quad \rightarrow \quad m_{\perp}=-\dfrac{1}{3}\\$

Next, find the midpoint of (4, 1) and (2, -5):

$Midpoint=\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{4+2}{2},\dfrac{1-5}{2}\bigg)\\\\\\.\qquad \qquad=\bigg(\dfrac{6}{2},\dfrac{-4}{2}\bigg)\\\\\\.\qquad \qquad=(3, -2)$

Lastly, input the perpendicular slope and the midpoint into the Point-Slope formula to find the equation of the line:

$y - y_1 = m_{\perp}(x - x_1)\\\\y - (-2) = -\dfrac{1}{3}(x - 3)\\\\y + 2=-\dfrac{1}{3}x +1\\\\y =-\dfrac{1}{3}x - 1\\$

6 0

2 years ago