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pashok25 [27]
3 years ago
9

Represent the following sentence as an algebraic expression, where "a number" is the

Mathematics
1 answer:
Vlad1618 [11]3 years ago
4 0
Answer:
2 (x+3)

twice : multiply by 2
sum of a number and 3 : x+3
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<img src="https://tex.z-dn.net/?f=The%20%5C%3A%20%20value%20%20%5C%3A%20of%20%20%5C%3A%20%20%5Csqrt%7B6%20%2B%20%20%5Csqrt%7B6%2
saul85 [17]

\large\underline{\sf{Solution-}}

<u>Let us assume that:</u>

\sf \longmapsto x =  \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ... \infty } } }

We can also write it as:

\sf \longmapsto x =  \sqrt{6 +  x }

Squaring both sides, we get:

\sf \longmapsto  {x}^{2}  =6 +  x

\sf \longmapsto  {x}^{2} - x - 6  =0

By splitting the middle term:

\sf \longmapsto  {x}^{2} - 3x + 2x - 6  =0

\sf \longmapsto x(x - 3) + 2(x - 3 ) =0

\sf \longmapsto (x+ 2)(x - 3 ) =0

<u>Therefore:</u>

\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}

\sf \longmapsto x =  - 2,3

<u>But x cannot be negative. </u>

\sf \longmapsto x = 3

Therefore, the value of the expression is 3.

\large\underline{\sf{Verification-}}

Given:

\sf\longmapsto x=3

We can also write it as:

\sf\longmapsto x = \sqrt{9}

\sf\longmapsto x = \sqrt{6+3}

\sf\longmapsto x = \sqrt{6 + \sqrt{9}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}

This pattern will continue.

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}

7 0
3 years ago
Por favor ayúdeme!!​
Inessa [10]

Answer:

CHEEEEEEEEESE

Step-by-step explanation:

3 0
4 years ago
PLEASE HELP!!!! Find the value of -7 - 8 - (-8) - 7. 0 14 -14 -30
miskamm [114]

-7 - 8 - (-8) - 7 × 0(14) - 14 - 30

(I think that's what that says)

Do PEMDAS

0(14) = 0

-7 × 0 = 0

You're left with -7 - 8 - (-8) - 14 - 30

Go from left to right

-7 - 8 = -15

-15 - (-8) = -7

-7 - 14 = -21

-21 - 30 = -51

7 0
3 years ago
Read 2 more answers
Please anyone help
serg [7]
5.46 to 3 sig figs means the range is {5.455,5.465} and 17.74 means the range {17.735,17.745}.
p=q²/r has a maximum value when q=5.465 and r=17.735 and a minimum value when 5.455 and r=17.745.
So the range of p is 1.6769 to 1.6840. When we have 2 decimal places we get p=1.68 which accommodates the maximum and minimum values of the range. So 2 decimal places is a suitable degree of accuracy, or we could say 3 significant figures.
7 0
3 years ago
Read 2 more answers
For the circle with equation (x-2)^2+(y+3)^2=9, answer each question. a) whats coords of the center? b) what are the radius and
MrRissso [65]

Answer:

<h2>a) center (2, -3)</h2><h2>b) radius r = 3</h2><h2>c) in the attachment</h2>

Step-by-step explanation:

The standard form of an equation of a circle:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have:

(x-2)^2+(y+3)^2=9\\\\(x-2)^2+(y-(-3))^2=3^2

a) center (2, -3)

b) radius r = 3

c) in the attachment

4 0
3 years ago
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