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saul85 [17]
3 years ago
8

N is the middle integer of three consecutive positive integers.

Mathematics
1 answer:
abruzzese [7]3 years ago
6 0

Answer :

\boxed{\textsf{ The final answer is \textbf{n}$^{\textbf{3}}$ .}}

Step-by-step explanation:

Its given that n is the middle out of the three consecutive integers . So ,

<u>The </u><u>last</u><u> </u><u>integer</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>:</u><u>-</u><u> </u>

\sf\implies Last \ Integer \ = \ n - 1

<u>The </u><u>next</u><u> </u><u>Integer</u><u> </u><u>will</u><u> be</u><u> </u><u>:</u><u>-</u>

\sf\implies Next \ Integer \ = \ n + 1

Now the Question says that the three integers are multipled to give a product . So that would be.

\sf\implies Product_{(three\ consecutive\ integers)}= (n-1)n(n+1) = (n^2-1)(n) = \pink{n^3-n}

Now thirdly it's given that n is added to the given integer . That would be ,

\sf\implies Adding\ n = \ n^3 - n + n = \pink{n^3}

Here - n and +n gets cancelled. So we are ultimately left out with n³.

Hence the final number is a cube of some number.

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Graph the image of this figure after a dilation with a scale factor of 13 centered at the point (4, −2) .
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From the given triangle figure;

Labelled the triangle as A , B and C.

The coordinates of this triangle ABC  are;

A = (1, 10) ,

B = (-2, 4)

C = (7, 4).

Given : Scale factor(k) = \frac{1}{3} and centered at point (4, -2).

The rule of dilation with k= \frac{1}{3} and center at point (4,-2) is:

(x, y) \rightarrow (\frac{1}{3}(x-4)+4 , \frac{1}{3}(y+2)-2)

(x, y) \rightarrow (\frac{1}{3}x-\frac{4}{3}+4 , \frac{1}{3}y+\frac{2}{3}-2)

or

(x, y) \rightarrow (\frac{1}{3}x+\frac{8}{3} , \frac{1}{3}y-\frac{4}{3})

then, the dilation of the given figure are;

A(1, 10) \rightarrow (\frac{1}{3}\cdot 1+\frac{8}{3} , \frac{1}{3} \cdot 10-\frac{4}{3}) = (\frac{1}{3}+\frac{8}{3} , \frac{10}{3}-\frac{4}{3}) =(\frac{9}{3} , \frac{6}{3}) = A'(3 , 2)

B(-2, 4) \rightarrow (\frac{1}{3}\cdot -2+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3}) =(-\frac{2}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3}) =(\frac{6}{3} , \frac{0}{3}) =B'(2 , 0)

C(7, 4) \rightarrow (\frac{1}{3}\cdot 7+\frac{8}{3} , \frac{1}{3} \cdot 4-\frac{4}{3}) =(\frac{7}{3}+\frac{8}{3} , \frac{4}{3}-\frac{4}{3}) =(\frac{15}{3} , \frac{0}{3}) = C'(5 , 0)

The coordinates of dilation images are:

A' = (3,2) , B' = (2, 0) and C' = (5, 0)

You can see the graph of the dilated image as shown below:

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