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jolli1 [7]
3 years ago
14

Circles c and c are similar state the translation rule and the scale factor of dilation

Mathematics
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

To find the scale factor for a dilation, we find the center point of dilation and measure the distance from this center point to a point on the preimage and also the distance from the center point to a point on the image.

Step-by-step explanation:

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Step-by-step explanation:

8 0
3 years ago
write the x and y coordinate (in the second and third column, respectively) of dilation of quadrilateral ABCD with vertices A(1,
gayaneshka [121]

Answer:

The coordinates of the dillated vertices are A'(x,y) = (2,2), B'(x,y) = (4,4), C'(x,y) = (8,2) and D'(x,y) = (4,-2).

Step-by-step explanation:

From Linear Algebra, we define dilation by the following equation:

P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)] (1)

Where:

O(x,y) - Center of dilation, dimensionless.

P(x,y) - Original point, dimensionless.

k - Scale factor, dimensionless.

P'(x,y) - Dilated point, dimensionless.

If we know that O(x,y) = (0, 0), k = 2, A(x,y) = (1,1), B(x,y) = (2,2), C(x,y) = (4,1) and D(x,y) = (2,-1), then the dilated points are, respectively:

Point A

A'(x,y) = O(x,y) + k\cdot [A(x,y)-O(x,y)] (2)

A'(x,y) = (0,0) + 2\cdot [(1,1)-(0,0)]

A'(x,y) = (2,2)

Point B

B'(x,y) = O(x,y) + k\cdot [B(x,y)-O(x,y)] (3)

B'(x,y) = (0,0) + 2\cdot [(2,2)-(0,0)]

B'(x,y) = (4,4)

Point C

C'(x,y) = O(x,y) + k\cdot [C(x,y)-O(x,y)]

C'(x,y) = (0,0) + 2\cdot [(4,1)-(0,0)]

C'(x,y) = (8,2)

Point D

D'(x,y) = O(x,y) + k\cdot [D(x,y)-O(x,y)]

D'(x,y) = (0,0) + 2\cdot [(2,-1)-(0,0)]

D'(x,y) = (4,-2)

The coordinates of the dillated vertices are A'(x,y) = (2,2), B'(x,y) = (4,4), C'(x,y) = (8,2) and D'(x,y) = (4,-2).

4 0
3 years ago
I'm stuck on this question, any ideas? thanks
True [87]

Answer:

\lim_{x \to 0} \frac{\frac{1}{6+x}-\frac{1}{6}}{x}=-1/36

Step-by-step explanation:

So we have the limit:

\lim_{x \to 0} \frac{\frac{1}{6+x}-\frac{1}{6}}{x}

Let's remove the fractions in the denominator by multiplying both layers by (6+x)(6). So:

\lim_{x \to 0} \frac{\frac{1}{6+x}-\frac{1}{6}}{x}\cdot (\frac{(6+x)(6)}{(6+x)(6)})

Distribute:

=\lim_{x \to 0} \frac{(6)-(6+x)}{x(6+x)(6)}

Simplify the numerator:

=\lim_{x \to 0} \frac{6-6-x}{x(6+x)(6)}\\=\lim_{x \to 0} \frac{-x}{x(6+x)(6)}

Both the numerator and the denominator have an x. Cancel:

=\lim_{x \to 0} \frac{-1}{(6+x)(6)}

Direct substitution:

= \frac{-1}{(6+0)(6)}

Simplify:

=-1/36

And that's our answer.

And we're done!

8 0
3 years ago
Read 2 more answers
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