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postnew [5]
4 years ago
15

If c is a positive integer, how does the graph of y=x+c compare to the graph of y=x?

Mathematics
1 answer:
Len [333]4 years ago
3 0

Answer:

It is shifted c units to the left  .

Step-by-step explanation:

In general, adding a constant c to the value of x shifts a graph c units to the left.

The graph of y = x + c is parallel to y = x,  the y-intercept becomes (0, c) and the x-intercept becomes (-c, 0).

In the diagram below, the red line is the graph of y = x, and the blue line is the graph of y = x + 1.

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Complete the equation of the line through (-8,-2) and (-4,6)<br> Use exact numbers.<br> y=
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3 years ago
Select each number that is between 535% and 6.01 x 10⁰.<br> 10 poin
Vikki [24]

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4 0
3 years ago
Estimate the square root to the nearest integer and tenth square root of 8
Sever21 [200]

Answer:

√

8

≈

3

Explanation:

Note that:

2

2

=

4

<

8

<

9

=

3

2

Hence the (positive) square root of  

8

is somewhere between  

2

and  

3

. Since  

8

is much closer to  

9

=

3

2

than  

4

=

2

2

, we can deduce that the closest integer to the square root is  

3

.

We can use this proximity of the square root of  

8

to  

3

to derive an efficient method for finding approximations.

Consider a quadratic with zeros  

3

+

√

8

and  

3

−

√

8

:

(

x

−

3

−

√

8

)

(

x

−

3

+

√

8

)

=

(

x

−

3

)

2

−

8

=

x

2

−

6

x

+

1

From this quadratic, we can define a sequence of integers recursively as follows:

⎧

⎪

⎨

⎪

⎩

a

0

=

0

a

1

=

1

a

n

+

2

=

6

a

n

+

1

−

a

n

The first few terms are:

0

,

1

,

6

,

35

,

204

,

1189

,

6930

,

...

The ratio between successive terms will tend very quickly towards  

3

+

√

8

.

So:

√

8

≈

6930

1189

−

3

=

3363

1189

≈

2.828427

8 0
3 years ago
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