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Nicely explained, Melody....!!!!

Thanks, DarkCalculus....!!!

3√8 x 6√5  =     {since we can multiply in any order }

3 * 6 * √8 * √5    =

18 * √40  =

18 * √4 * √10  =

18 * 2 * √10  =

36√10

no problem 

thanks

I really do not understand your question Manuel.

I do not understand what you mean here 

Radical 5 and Radical 4 with 3 index

AND i do not understand what you mean by 'the same order'

If you know what you mean, perhaps you can explain better, or upload a photo of the actual question perhaps. :/

1.A telegraph has x arms and each arm is capable of (x-1) distinct positions , including the position of rest. The total no. of signals that can be made is?

Rosala, This is just      x(x-1)

2. How many natural numbers are there fro 1 to 1000 which have none of their digits repeated?

i've tried this question a no. of times but it just doesnt see to be clicked in my mind. i would be really appreciate if anyone can explain it to me in detail.

There are 9 one digit ones

There are no four digit ones

So how many 2 digit ones are there.

The tens digit can be 1 to 9 that is 9 choices, now you used one digit but you can use the zeros so there are 9 possible digits left for the units digit

So that is 9*9=81

AND how many three digit ones are then  9*9*8 = 648

Altogether there are  9+81+648 = 738

3. Number of different natural numbers which are smaller than two hundred million and using only the digits 1 or 2 is :

less than      200,000,000   only containing the digits 1 and 2

1 digit     2

2 digit     2*2=2^2

3 digit      2^3

4 digit       2^4

5 digit      2^5

6 digit      2^6

7 digit      2^7

8 digit      2^8

9 digit      2^8       The biggest value digit must be 1

So what do we get when we add these up

\(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^8\\ =(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)+2^8\\ \qquad \text{The brackets is the sum of a GP, a=2 and r=2\;\;n=8}\\ =\frac{a(r^n-1)}{r-1}\;\;\;+2^8\\ =\frac{2(2^8-1)}{2-1}\;\;\;+2^8\\ =2^9-2+2^8\\ =2^9+2^8-2\\ =2^8(2+1)-2\\ =3\times 2^8-2 \)

i. (3).2^8 - 2 

ii. (3).2^8 - 1 

iii. 2 (2^9 - 1) 

iv) None 

I have a few more question like the questions above, and i just dont seem to get the correct answers. 

i have tried the questions above many times but i just cant understand a few things , i would be glad if anyone could explain to me in detail. Thank you very much. 

i really appreciate the time you'd take out for my answers! 

Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0. Find \frac{1}{2a} + \frac{1}{2b}.

\(\text{Let }\alpha\;\; and \;\ \beta\;\; \text{be the solutions of the quadratic equation }\\ 2x^2 - 8x + 7 = 0. \;\;\;Find \;\;\ \frac{1}{2a} + \frac{1}{2b}\\ \frac{1}{2\alpha} + \frac{1}{2\beta}=\frac{\beta + \alpha}{2\alpha\beta} \)

Now you can do this the long way and work out what the roots are but I expect you are supposed to know this:

\(\boxed{If \;\;ax^2+bx+c=0 \text{ and the roots are }\alpha \;\;and \;\; \beta\;\;then\\ \alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\}~\\ \)

So for your question

\(\alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\ \alpha + \beta = \frac{8}{2}\;\;and \;\; \alpha \beta = \frac{7}{2}\\\)

\(\frac{1}{2\alpha} + \frac{1}{2\beta}\\=\frac{\beta + \alpha}{2\alpha\beta}\\ =\frac{8}{2}\div\frac{2\times7}{2}\\ =4\div7\\ =\frac{4}{7}\)

Thanks Rosala and Manuel,

A very big thank you to both of you.  

I think I speak for all the answerers here :)

AND

Rosala, what beautiful flowers and chocolate cake and I get a bit of both. 

Thanks  

In the diagram below, we have DE = 2EC and AB = DC = 20. Find the length of FG.

\(\text{Let $DC = 20$ } \\ \text{Let $DE = \dfrac23\cdot 20$ } \\ \text{Let $EC = \dfrac13\cdot 20$ } \\ \text{Let $FG = x + EC $ or $x=FG-EC $ }\\ \text{Let $BG = p $ } \\ \text{Let $GC = q $ } \)

1. intercept theorem

\(\begin{array}{|rcll|} \hline \dfrac{q}{x} &=& \dfrac{p}{DE-x} \quad & \quad DE= \dfrac23\cdot 20 \\\\ \dfrac{q}{x} &=& \dfrac{p}{\dfrac23\cdot 20-x} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{40-3x}{3x} \qquad (1)} \\ \hline \end{array} \)

2. intercept theorem

\(\begin{array}{|rcll|} \hline \dfrac{q}{DE-x} &=& \dfrac{p}{x+EC} \quad & \quad DE= \dfrac23\cdot 20 \qquad EC=\dfrac13\cdot 20 \\\\ \dfrac{q}{ \dfrac23\cdot 20-x} &=& \dfrac{p}{x+\dfrac13\cdot 20} \\\\ \mathbf{\dfrac{p}{q}} & \mathbf{=} & \mathbf{\dfrac{3x+20}{40-3x} \qquad (2)} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(2): & \mathbf{\dfrac{p}{q}} = \mathbf{\dfrac{40-3x}{3x} }&\mathbf{=}& \mathbf{\dfrac{3x+20}{40-3x}} \\\\ & \dfrac{40-3x}{3x} & = & \dfrac{3x+20}{40-3x} \\\\ & (40-3x)^2 & = & 3x(3x+20) \\\\ & 1600-240x+\not{9x^2} & = & \not{9x^2} +60x \\\\ & 1600-240x & = & 60x \\\\ & 300x &=& 1600 \quad & | \quad : 300 \\\\ & x &=& \dfrac{1600}{300} \\\\ & x &=& \dfrac{16}{3} \\\\ & FG &=& x + EC \qquad EC=\dfrac{20}{3} \\\\ & &=& \dfrac{16}{3} + \dfrac{20}{3} \\\\ & &=& \dfrac{36}{3} \\\\ & \mathbf{FG} &\mathbf{=} & \mathbf{12} \\ \hline \end{array}\)

The interest over 3 years is $420.24 which is $140.08 per year

therefore the rate per year is

140.08/7004 = .02 and this is 2%

The answer of 121 is close to my calculation of 117 if that's any help ?  

An alternative method you can use especially if you are familiar with factoring quadratic trinomials is the following 

moving on from x² + x = 30 you move the 30 across to the left hand side 

x²  +  x  -  30 =  0

then  (x+6)(x-5) = 0

conclusion : if 2 factors multiply to give 0 then either can be 0

ie x+6 =0  or  x-5=0  

therefore  x=-6 or x=5

Thank You Cphill

Yes, I am Also Grateful for all your Help, Cphill, Melody, Hectitar (to name a few). But I am thankful to all of you, Moderators and Users too.  You Guys Make School Life worth living!!! . Thank You!!!

You Guys are there answering our questions and Explaining them patiently. Thank You all

Three Cheers for The Moderators and Users!! (Even some guests too!!)

Hurrah, Hurrah, Hurrah!!!

Thank you so much. I am so grateful!!😊

\(4x^2 - 31x = 8\)

                                      Subtract  8  from both sides of the equation.

\(4x^2 - 31x - 8 = 0\)

Now we can solve this quadratic equation using the quadratic formula.

The quadratic formula says...

If     \(ax^2+bx+c=0\)    then    \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

So...

If    \(4x^2 - 31x - 8 = 0\)    then    \(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)}\)

\(x = {-(-31) \pm \sqrt{(-31)^2-4(4)(-8)} \over 2(4)} \\~\\ x = {31 \pm \sqrt{961+128} \over 8} \\~\\ x = {31 \pm \sqrt{1089} \over 8} \\~\\ x = {31 \pm 33 \over 8} \\~\\ \begin{array} \ x=\frac{31+33}{8}\qquad&\text{or}&\qquad x=\frac{31-33}{8} \\~\\ x=\frac{64}{8}&\text{or}&\qquad x=\frac{-2}{8} \\~\\ x=8&\text{or}&\qquad x=-\frac{1}{4} \end{array}\)

Thanks, guest.....correction made  !!!

Suppose that we have an equation y = ax^2 + bx + c whose graph is a parabola with vertex (3,2), vertical axis of symmetry, and contains the point (1,0).

What is (a , b, c)?

In the vertex form, we have

y  = a(x - h) +  k   ⇒   (h , k)  =  (3,2)   ....  so.....

0  =  a(1 - 3)^2 + 2       subtract 2 from both sides

-2  = a (-2)^2

-2  = 4a        divide both sides by 4

-2/4  =  a  =  - 1/2

So we have

y = (-1/2)(x -3)^2 + 2

y = (-1/2)(x^2 - 6x + 9) + 2

y =  (-1/2)x^2 + 3x - 9/2 + 2

y  = (-1/2)x + 3x - 5/2   ⇒   (a, b, c)  =  (-1/2, 3, -5/2)

Here's the graph  :  https://www.desmos.com/calculator/wxrh0uke2u

Thank you so much! I am failing in math and this means so much!!!

A typo of 139 should be 169.

We already know that x + y   = 3  .....  do you mean x  - y  ???

x^4 = y^4 + 18√3,  x^2 + y^2  =  6,  x + y  =  3

Using the first equation

x^4 - y^4  = 18√3      factor the left side

(x^2  - y^2) (x^2 + y^2)  =18√3        sub in the second equation

(x^2 - y^2) (6)  =  18√3      divide both sides  by 6

(x^2  - y^2)  = 3√3    factor again  

( x - y) (x + y)  = 3√3  sub in the third equation

(x - y)  (3)  =  3√3    divide both sides by 3

x  - y   =  √3

Solve for x:

2 x^2 + 2 x = 60

Divide both sides by 2:

x^2 + x = 30

Add 1/4 to both sides:

x^2 + x + 1/4 = 121/4

Write the left hand side as a square:

(x + 1/2)^2 = 121/4

Take the square root of both sides:

x + 1/2 = 11/2 or x + 1/2 = -11/2

Subtract 1/2 from both sides:

x = 5 or x + 1/2 = -11/2

Subtract 1/2 from both sides:

x = 5         or          x = -6

15^6  - 7^6  =  {factor using  difference/sum of cubes}

(15 - 7)(15^2 + 15*7 + 7^2)(15 + 7) (15^2 - 15*7 + 49)  =

(15 - 7) (15+7)(15^2 - 15*7 + 49)(15^2 + 15*7 + 49) =

  (8)(22)(169)(379)

(2^3)(2*11)(13* 13)(379)

(2^4)(11)(13^2)(379)    is the largest prime factor