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iren [92.7K]
3 years ago
5

finding the distance from (–6, 2) to the origin. Use (0, 0) as (x1, y1). d = StartRoot (x 2 minus x 1) squared + (y 2 minus y 1)

squared EndRoot What is the difference of the x-coordinates, (x2–x1)? What is the difference of the y-coordinates, (y2–y1)? What is the distance from (–6, 2) to the origin?
Mathematics
1 answer:
Naddik [55]3 years ago
4 0

Answer:

2√10

Step-by-step explanation:

Given the coordinates (-6, 2) and (0,0)

We are to find the distance between the coordinates. Using the distance formula;

d = √(x2-x1)²+(y2-y1)²

d = √(0-2)²+(0+6)²

d = √(-2)²+(6)²

d = √4+36

d = √40

d = √4*10

d = 2√10

Hence the required distance is 2√10

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In the figure, triangle CAE is an enlargement of triangle CBD with scale factor of 4/3. The area of the smaller triangle is 9cm2
balandron [24]

Answer:

16 cm^2

Step-by-step explanation:

Given

\triangle CAE -- Bigger Triangle

\triangle CBD -- Smaller Triangle

k = \frac{4}{3} --- Scale factor

Area of CBD = 9

Required

Determine the area of CAE

The area of triangle CBD is:

A_1 = \frac{1}{2}bh

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The area of CAE is:

A_2 = \frac{1}{2}BH

Where:

B = \frac{4}{3}b and

H = \frac{4}{3}h

The above values is the dimension of the larger triangle (after dilation).

So, we have:

A_2 = \frac{1}{2}*\frac{4}{3}b * \frac{4}{3} * h

A_2 = \frac{1}{2}*\frac{4}{3} * \frac{4}{3} *b* h

A_2 = \frac{1}{2}*\frac{16}{9}  *b* h

Re-order

A_2 = \frac{16}{9}*\frac{1}{2}* b* h

A_2 = \frac{16}{9}*\frac{1}{2}bh

Recall that:

\frac{1}{2}bh = 9

A_2 = \frac{16}{9}*9

A_2 = 16

Hence, the area is 16 cm^2

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