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Marta_Voda [28]
1 year ago
13

triangle ABC is dilated by a scale factor of four to form triangle A'B'C'. what are the coordinates of vertices A', B', and C'

Mathematics
1 answer:
netineya [11]1 year ago
3 0

The question did not show the original vertices of A, B, and C.

Generally, for a triangle ABC that is dilated by a scale factor of four to form triangle A'B'C', the coordinates of vertices A', B', C' are:

For A(x, y)

The coordinate of vertex A' will be A' (4x, 4y)

For B(x, y)

The coordinate of vertex B' will be B' (4x, 4y)

For C(x, y)

The coordinate of vertec C' will be C' (4x, 4y)

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4. Diego is thinking of two positive numbers. He says, "If we triple the first number and
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3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are  (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).

Step-by-step explanation:

 Let the First positive number be <em>x</em> and second positive number be <em>y.</em>

Triple of first number = 3<em>x</em>

double of second number = 2<em>y</em>

According to question,

3<em>x </em>+ 2<em>y</em> = 34

Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34

So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34

Now,  3<em>x </em>+ 2<em>y</em> = 34

3<em>x </em>= 34 - 2<em>y</em>

Let x = 10 and by substituting its value in above expression,

3 \times 10 = 34 - 2y

30 = 34 - 2y

2y = 34 - 30

2y = 4

y = 2

Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)

In the same way put x = 4 then,

3<em>x </em>= 34 - 2<em>y</em>

3 \times 4 = 34 - 2y

2y = 34 - 12

2y = 22

y = 11

Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)

Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and  (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.  

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