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nignag [31]
1 year ago
13

Can someone help me with this question in so confused

Mathematics
1 answer:
Svetlanka [38]1 year ago
3 0

Answer:

D(1, 2) → D'(2, 7)

E(-3, -5) → E'(-10, 0)

F(4, -1) → F'(11, 4)

Step-by-step explanation:

D(1, 2) → D'   ________

Translate image (3x - 1, y + 5)

(1, 2), x = 1 and y = 2

3x - 1 = 3(1) - 1 = 3 - 1 = 2

y + 5 = (2) + 5 = 7

E(-3, -5) → E' ________

(-3, -5), x = -3 and y = -5

3x - 1 = 3(-3) - 1 = -9 - 1 = -10

y + 5 = (-5) + 5 = 0

F(4, -1) → F'   ________

(4, -1), x = 4 and y = -1

3x - 1 = 3(4) - 1 = 12 - 1 = 11

y + 5 = (-1) + 5 = 4

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Answer:

1, 27, 3, 9,

Step-by-step explanation:

To find the factors of 27 you need to find which whole number times another whole number = 27

We know that 1*27 = 27

We also know that 3*9 = 27

The factors are all the numbers that are used in the multiplication.

In this example the numbers are: 1, 3, 9 and 27

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GIVEN:

5(u - 1)( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1)

remember:

\sqrt{u} = {u}^{ \frac{1}{2} }

And

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SOLVE:

start by multiplying the factors:

5( ({u}^{6} + {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u ) - ( {u}^{5} + {u}^{4} + {u}^{3} + {u}^{2} + u + 1))

simplify by combing like terms. Most terms subtract off, leaving:

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This can be factored, but it is not a perfect square, which is really what we need to take the square root.

5( {u}^{3} - 1)( {u}^{3} + 1)

I'm not exactly sure what form they want the answer in...

so taking the square root:

\sqrt{5( {u}^{6} - 1) } = {(5( {u}^{6} - 1))}^{ \frac{1}{2} }

so my best answer is:

{5}^{ \frac{1}{2} } \times {( {u}^{6} - 1)}^{ \frac{1}{2} }
or the more factored form:

{5}^{ \frac{1}{2} } { ({u}^{3} - 1)}^{ \frac{1}{2} } { ({u}^{3} + 1 )}^{ \frac{1}{2} }

I'm not sure how else to solve it. Taking the square root doesn't work out super well, so I left it in the most simple form I could.

sorry for not coming to a definitive answer!
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