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solong [7]
3 years ago
7

George has a triangle-shaped garden in his backyard. He drew a model of this garden on a coordinate grid with vertices A(4, 2),

B(2, 4), and C(6, 4). He wants to create another, similar-shaped garden, A′B′C′, by dilating triangle ABC by a scale factor of 0.5. What are the coordinates of triangle A′B′C′?
A.
A′(2, 2), B′(1, 2), C′(3, 2)
B.
A′(2, 1), B′(1, 2), C′(3, 2)
C.
A′(8, 4), B′(4, 8), C′(12, 8)
D.
A′(2, 1), B′(1, 2), C′(6, 4)
Mathematics
2 answers:
Nana76 [90]3 years ago
8 0

Answer:

B

Step-by-step explanation:

you divided each cordinate by 2 so (4,2)/2 is (2,1) .

(2,4)/2 = (1,2)

6,2/2 =(3,2)

Otrada [13]3 years ago
3 0

Answer:

b). A' (2,1) , B' (1,2), C' (3,2)

Step-by-step explanation:

Scale factor 0.5 = 1:2

Scale factor = \frac{1}{2} = 0.5

A'= \frac{A}{2} =\frac{(4,2)}{2} =(2,1)\\ B'= \frac{B}{2} =\frac{(2,4)}{2} =(1,2)\\C'= \frac{C}{2} =\frac{(6,4)}{2} =(3,2)\\

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Salsk061 [2.6K]

Answer:

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Step-by-step explanation:

What value is close to 28 that is a perfect cube...27 which equals 3^3.

So let's find the tangent line to the curve y=cubert(x) at x=27.

We will use this equation to approximate what happens at x=28.

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Now differentiate

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Simplify

y'=(1/3) x^(-2/3) or (1)/(3x^[2/3])

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So at x=27, we have y=cubert(27)=3. We used our equation y=cubert(x) here.

So we want to find the equation of the line that contains point (27,3) and has slope 1/27.

Point-slope form is

y-y1=m(x-x1)

Plug in our values

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Add 3 on both sides

y=3+(1/27)(x-27)

We will use this linear equation to approximate cubert(28) by replacing x with 28.

y=3+(1/27)(28-27)

y=3+(1/27)(1)

y=3+1/27

You can write that as a mix fraction if you want.

This value is than 3 but super close to 3 since 1/27 is close to 0.

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Decimal approximation: 3.037

Cubert of 28 when smashed into calculator as is gives approximately 3.0366 which is pretty close to our approximation.

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