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Lerok [7]
3 years ago
11

A polygon has vertices A(1,1), B(1, 4), C(4, 4). This figure is dilated by a factor of 3 and center of dilation (-1, 4). Find th

e coordinates of the polygon after dilation.
Mathematics
1 answer:
Art [367]3 years ago
5 0

Answer:

Step-by-step explanation:

New coordinate = [old coordinate - dilation center]* scale factor + dilation center

So,

For A(1,1)

[(1,1)-(-1,4)]*3+(-1,4)\\(2,-3)*3+(-1,4)\\(6,-9)+(-1,4)\\(5,-5)

For B(1,4)

[(1,4)-(-1,4)]*3+(-1,4)\\(2,0)*3+(-1,4)\\(6,0)+(-1,4)\\(5,4)

For C(4,4)

[(4,4)-(-1,4)]*3+(-1,4)\\(5,0)*3+(-1,4)\\(15,0)+(-1,4)\\(14,0)

Therefore these (5,-5) , (5,4) and (14,0) are the new coordinates of polygon after dilation.

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AleksandrR [38]

Answer:

a = 145

b = 35

c = 145

d = 70

e = 70

f = 110

g = 55

h = 125

i = 55

j = 50

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m = 50

n = 60

o = 70

p = 23

q = 89

r = 68

s = 157

t = 112

u = 48

v = 132

w = 132

x = 48

y = 48

z = 132

A = 48

B = 94

C = 86

D = 94

E = 47

F = 133

Step-by-step explanation:

I'm fairly certain these are all correct... like 82% sure

3 0
3 years ago
How do I do it and show my work along with it?
Rudiy27
Be kind and say please,that is being rude when you don't say it. Well, how much times does 6 go into 34? 5 times right. So then put 30 under the 34. You get 4. So then bring down the 5. How much times does 6 go into 45. 7 right. So then put 42 under 45. You get 3. Now add a decimal point after the 5 and after the 7. Add zero after the 5 . Bring down the zero. How much times does 6 go into 30? 5 times, so you put 30 under 30 and get zero. If you have any more questions,please leave them below in the comments.
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3 years ago
What percent is 56 of 200​
GrogVix [38]

Answer:

28%

Step-by-step explanation:

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O is the centre of the circle and ABC and EDC are tangents to the circle
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Answer:

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7 0
3 years ago
Multiply. Check picture.
VashaNatasha [74]

The answer is 3x^4-13x^3-x^2-11x+6.

Solution:

Use algebraic identity: a^m\times a^n=a^{m+n}

For example: x^2\times x=x^{2+1}=x^3

Given expression (x^2-5x+2) and (3x^2+2x+3).

To multiply these equations.

(x^2-5x+2)\times(3x^2+2x+3)

             =x^2(3x^2+2x+3)-5x(3x^2+2x+3)+2(3x^2+2x+3)

             =(3x^4+2x^3+3x^2)+(-15x^3-10x^2-15x)+(6x^2+4x+6)

             =3x^4+2x^3+3x^2-15x^3-10x^2-15x+6x^2+4x+6

Combine like terms together.

             =3x^4+(2x^3-15x^3)+(3x^2-10x^2+6x^2)-15x+4x+6

             =3x^4-13x^3-x^2-11x+6

(x^2-5x+2)\times(3x^2+2x+3)=3x^4-13x^3-x^2-11x+6

Hence the answer is 3x^4-13x^3-x^2-11x+6.

3 0
3 years ago
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