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(x^3-5x^2+7x-12)÷(x-4)

             x^2   -  x    +  3

x - 4  [  x^3  -5x^2  + 7x  -  12  ]

            x^3 - 4x^2

          ______________________

                  - x^2  +  7x

                  - x^2  + 4x

                __________________

                            3x   -   12

                            3x   -   12

                          __________

The result is in red

2)  Here's the graph : https://www.desmos.com/calculator/hz0nsgeb2n

The x intercept   is  (1, 0)

The y intercept is (0, - 3)

1)  (a)  There will be as many roots of a polynomial as its degree  [  some may be repeated roots ]

So....the number of roots here    = 2    since this is a second degree polynomial

(b)   2x^2 + 4x + 7  = 0     subtract 7 from both sides

2x^2 + 4x  =  - 7      factor out the 2 on the left

2 ( x^2 + 2x )   =  - 7    divide both sides by 2

x^2 + 2x  =  -7/2  

Take 1/2 of 2  = 1...square it = 1....add it to both sides

x^2 + 2x + 1   =  -7/2 + 1      factor the left side, simplify the right

(x + 1)^2  = -5/2     take both roots

x + 1  =  ±√ [ - 5/2]  =  ±i √ 5 / √2  =   ±i √10 /2

x + 1  = ±i√10 / 2    subtract 1 from both sides

x =  -1 ±i√10 / 2

x =  [  -2  ± i √10 ]  / 2

Thanks, EP....I sorta' expected that THAT was the question that was meant to be asked, but I wasn't sure....

In  the form    ax^n  + bx^m +.......  + c  =  0

The possible rational  roots  [ rational meaning that we can express a number as a fraction]  are all the possible factors  of  "c"   over all the possible factors of "a"

So we have

9x^3  + 14x^2 - x + 18   =  0

So...all the possible  factors of  c = 18  are ±18, ±9, ±6, ±3, ±2, ±1

And all the possible factors of  a  = 9   are  ±9, ±3, ±1

So....in your answer...all the possible rational roots  are    ±1/3  and ±3/1

Note that ±1/2  isn't possible because ±2  isn't a factor of  9

And  ±1/ 18  isn't possibe because ±18  isn't a factor of 9

Note something, PhoenixForever......this does not mean that these are necessarily the roots of the equation....they are just possible roots...!!!

Here's a resource I always like, Purplemath....it explains this in more detail than I can :

https://www.purplemath.com/modules/rtnlroot.htm

Chris has given you GA       G'A'  will be of proportionate length

14/35  = G'A' / 30

G'A' = 30  14/35 = 12 inches

Let x  be the length of GA

So.....the perimeter of the adult model is the sum of all the side lengths....so

56 + 56 + 35 +  x   =  177    simplify

147   +  x  =   177      subtract  147 from both sides

x  =  30   =    length of GA

Sorry, I just don't understand what you're trying to say

No  identity matrix  will included in PAQ and all other will have 0 exp. Will be 3 lines and 4 and will have identity matrix 1x1 or 2x2 or 3x3 and all other its 0 

So you're saying that PAQ is an identity matrix with an unknown number of columns?

SO it is like you have 6 cards      One black and 5 red

you draw a red, then what is the chance of the next draw is red.....?

   there are 5 cards left    4 are red and one is black

       4/5           is that how to view this ?

its an identity matrix with unknown lines and columns

What is Irxr?

You are very welcome.   

I am always keen to share my knowledge, and my enthusiasm for mathematics, with those that really want to learn. 

In fact I thrive on the opportunity!

The other regular answerers feel this way too.   

Just to verify that the above is true, I tested the first 500 numbers! As you can see, after n=6 the pattern repeats every 13  numbers after that. You can also see that GCD is always the same, or 13.

a=listfor(n, 1, 500, GCD(n + 7, 2*n + 1))

(1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13)

you've been very kind...thank you,...have a blessed day!!

\(\sqrt{900}*2=60, \sqrt{900}=30\)

pls solve

\(x^2=64=x=8\)

If you want to check you can put your functions into Desmos.

YES that appears to be correct.   

Thank you Melody, may I please just conclude:

This is WRONG:

\(tan(\theta-180)\)

\(=tan(-(\theta+180))\)

\(=-tan(\theta+180) \)

\(=-tan(180+\theta)\)

\(=-tan\theta\)

And this is RIGHT:

\(tan(\theta-180)\)

\(=tan(-(-\theta+180))\)

\(=-tan(-\theta+180) \)

\(=-tan(180-\theta)\)

\(=tan\theta\)

Thank you sooo much!!!

Prove that  \[1{n\choose 1} + 2{n\choose 2} + 3{n \choose 3} + \cdots +  n{n\choose n} = n2^{n-1}\] 

for all positive integers n.

\(\begin{array}{|rcll|} \hline && \mathbf{1\dbinom n1 + 2\dbinom n2 + 3\dbinom n3 + 4\dbinom n4 + \ldots + n\dbinom nn} \\\\ &=& 1\cdot\dfrac{n!}{(n-1)!\cdot 1} + 2\cdot\dfrac{n!}{(n-2)!\cdot 1\cdot 2} + 3\cdot\dfrac{n!}{(n-3)!\cdot 1\cdot 2\cdot 3} \\ &&+ 4\cdot\dfrac{n!}{(n-4)!\cdot 1\cdot 2\cdot 3\cdot 4} + \ldots + n\cdot\dfrac{n!}{(n-n)!\cdot 1\cdot 2\cdot \ldots \cdot(n-1)\cdot n } \\\\ &=& \dfrac{n!}{(n-1)!\cdot 0!} + \dfrac{n!}{(n-2)!\cdot 1!} + \dfrac{n!}{(n-3)!\cdot 2!} + \dfrac{n!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n!}{(n-n)!\cdot(n-1)! } \\\\ &&\boxed{n!=n\cdot(n-1)!} \\\\ &=& \dfrac{n\cdot(n-1)!}{(n-1)!\cdot 0!} + \dfrac{n\cdot(n-1)!}{(n-2)!\cdot 1!} + \dfrac{n\cdot(n-1)!}{(n-3)!\cdot 2!} + \dfrac{n\cdot(n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{n\cdot(n-1)!}{(n-n)!\cdot(n-1)! } \\\\ &=& n\cdot \Big[ \dfrac{ (n-1)!}{(n-1)!\cdot 0!} + \dfrac{ (n-1)!}{(n-2)!\cdot 1!} + \dfrac{ (n-1)!}{(n-3)!\cdot 2!} + \dfrac{ (n-1)!}{(n-4)!\cdot 3!} \\ && +\ldots + \dfrac{ (n-1)!}{(n-n)!\cdot(n-1)! } \Big] \\\\ &=& n\cdot \Big[ \underbrace{\dbinom{n-1}{0}+\dbinom{n-1}{1}+\dbinom{n-1}{2}+\dbinom{n-1}{3} +\ldots + \dbinom{n-1}{n-1}}_{=2^{n-1}} \Big] \\\\ &\mathbf{=}& \mathbf{n\cdot 2^{n-1}} \\ \hline \end{array}\)

I write my my maths in LaTex.

See the box in the ribbon above.  [I mean when you are creating a post]

Click into that and a box comes up in which you can input your latex coding.

You can start with very simple things and learn from there. Lots of people here will be happy to help you.

Latex is code that allows you to write mathematics things on the internet (it is a lot more than that really but that is what we use it for.)

anyway, open the LaTex input box and type in

tan(\theta-180)    then press "ok"

and this is what you weill get 

\(tan\theta\)

theta by itself is just coded as \theta 

Give it a go, LaTex coding is fun :)

Hi Melody,

just as I had suspected, The book is wrong for they indicate that;

tan(@-180) = tan(-(@+180))..

I thought that was incorrect. Thank you soo much for verifying for me....

just..how do you get the theta sign?.. 

Hi Juriemagic,

Not both are true but i expect that is because you made a typo.

Think about the cyclic nature of the tan function. It repeats every 180 degrees

If you consider it works on the unit circle, that should help.

\(tan\theta = tan(\theta\pm180k) \qquad \text{where k is an integer}\\ so\\ tan\theta = tan (\theta -180)=tan(-(-\theta+180))\\ also\\ tan\theta = tan(\theta+180)=tan(-(-\theta-180))\\ \\~\\ HOWEVER\\ tan \theta =tan(\theta-180)\ne tan(-(\theta+180))\\ but\\ tan \theta = -tan(-(\theta+180))\\\)

Here are some of those graphs