Point of concurrency help!! second question

Estimate the square root to the nearest integer and tenth square root of 8

Sever21 [200]

Answer:

√

8

≈

3

Explanation:

Note that:

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

, 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

Ln x-ln(x-5)=ln2<br> Help in this problem please having problems on my homework

Eddi Din [679]

17 because you need to multiply

3 0

3 years ago

Find y intercepte , x intercepte

Reptile [31]

Do u have an attachment maybe??

8 0

3 years ago

What is twice 230? Write using exponents. Explain your reasoning.

Zielflug [23.3K]

2 x 230 = 231
I’m not 100% sure if this is correct?? Hope it helps though

6 0

3 years ago

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

<u>Transverse axis</u>

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

<u>The vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

<u>The co-vertices</u>

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola