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No there is not enough information. 

The average rate of change is the gradient of the line joining the 2 end points.

Gradient is the change in f(x) divided by change in x.

What is the change in f(x) between  x=5 and x=9 ?  

What is the change in x between  x=5 and x=9 ?  

So what is the average rate of change?

Please no one answer over me.   I am trying to teach!

Thanks Omi for that beautifully presented answer but we (many of us anyway) are trying to get the askers to do some thinking for themselves.

This will increase learning and decrease cheating. 

That is why I very purposefully gave a part answer. 

Mmm

A portion of the graph of y = f(x) is shown in red below, where f(x) is a quadratic function. The distance between grid lines is 1 unit. What is the sum of all distinct numbers x such that f(f(f(x)))=-3 ?

f(0) = -3     and     f(-4)= -3

So now I need  f(x)=0 and f(x)=-4

f(2)=0,    f(-6)=0            f(-2)=-4

So now I need

f(x)=2,      f(x)=-6   and   f(x)=-2

f(2.8)=2,     f(-6.8)=2    and     none,    and   f(0.8)=-2      f(-4.8)=-2

f(2.8)=2

f(f(2.8)=f(2)=0

f(f(f(2.8))=f(f(2))=f(0)=-3     

same for the others 

so

the possible x values are 2.8,  -6.8,  0.8 and -4.8      (all approximate)

sum = -8

Guest, the content of your answer may well be better than mine but it is so blocked together that it would be very difficult for people to decipher.

Presentation is important.

b.) Explain what your answer tells you about x + 1 as a factor.

It tells you that when the binomial x + 1 is divided with j(x) then is it divided equally with a remainder of 0 so it is a factor.

c.) Algebraically find the remaining zeros of j(x).

Nice presentation GameMaster, I am impressed with your new style of delivery :)

You know that x+1 is a factor so if you divide  j(x) by x-1 you will find at least one more factor.

I have done it to check that it works nicely but I want you to the the division for yourself.

I just used normal algebraic long division but i suppose you could use synthetic division if you know that method.

Anyway, after you do that it is relatively easy to break the new factor into 3 to get all the  factors.

Once you have the factors you can get the zeros very easily.

Give it a go and if you get stuck then explain to us what your problem is  

\(\text{let the number be }abc\\ 49a+7b+c = 81c+9b+a\\ 48a-2b=80c\\ c = \dfrac{24a-b}{40}\)

I'll let you figure out what the valid numbers are using the equation for c above.

Remember digits can only range from 0-6.

And of course digits have to be whole numbers.

40 boys and 28 girls stand in a circle, hand in hand, all facing inward. Exactly 18 boys give their right hand to a girl. How many boys give their left hand to a girl?

If a boy gives his right hand to a girl then looking from the middle of the circle the girl is on the left.     

So from the centre of the cirle the pairs look like    GB (clockwise)

 So we have  18 pairs of   (GB)    these pairs must be kept together.         [What I need to determine is how many  BG there will be]

Now there are 22 boys and 10 girls not accounted for although it would actually make no difference at all how many more children that there were.These children can be slotted in anywhere in the circle SO LONG AS for each slot postion the boys go first and the girls go second ( BG clockwise). It does not matter how many boys or girls get slotted in between each pair as long as they all get slotted in.  

Between each of the original  GB couples, who can be inbetween

GB                           nothing                           GB                BG will be together  (once)        

GB                          all boys                            GB                BG will be to together once at end

GB                           all girls                            GB                 BG will be together once at the beginning

GB        some boys followed by some girls     GB                 BG will be together once in the middle.

So no matter where the extra kids are sloted in (so long as the boys first then girls after rule is followed) there will be exactly ONE BG combination between each of the original pairs. HENCE if there are exactly 18 GB combinations there will also be exactly 18 BG combinations.

i.e.

If exactly 18 boys give their right hand to a girl THEN exactly 18 boys will give their left hand to a girl.  

Ther fact that there is 22 extra boys and 10 extra girls (above the 18 original pairs) is irrelevant.

CREDITS:

\(\dfrac{4x}{(x-5)(x-3)^2} = \dfrac{A(x-3)^2+B(x-5)(x-3)+C(x-3)}{(x-5)(x-3)^2}\)

\(4x = A(x^2-6x+9) + B(x^2-8x+15)+C(x-3)\\ 4x = x^2(A + B) + x(-6A-8B+C)+(9A+15B-3C)\)

\(A+B=0\\ (-6A-8B+C)=4\\ (9A+15B-3C) = 0\)

\(\text{I leave you to solve for }A,B,C\)

\((2^3)^n=2^1\\ 2^{3n}=2^1\\ 3n=1\\ n=\dfrac 1 3\)

\(\tan(60^\circ)=\dfrac{top}{50}\\ \tan(40^\circ)=\dfrac{base}{50}\\ top-base =50 \tan(60^\circ)-50\tan(40^\circ)\\ \text{I leave you to complete it}\)

\((3m)^2 + 12 = 3(m^2 + 12)\\ 9m^2 + 12 = 3m^2 + 36\\ 6m^2-24=0\\ m^2 - 4 = 0\\ \text{I leave you to finish}\)

e=2.718281828...

It's a transcendental number like Pi(π)=3.14159265359...

First, we note that there are two points on the graph whose y-coordinates are -3. These are (-4,-3) and (0,-3). Therefore, if f(f(f(x)))=-3, then f(f(x)) equals -4 or 0. There are three points on the graph whose y-coordinates are -4 or 0. These are (-2,-4), (-6,0), and (2,0). Therefore, if f(f(x)) is -4 or 0, then f(x) equals -2, -6, or 2. There are four points on the graph whose y-coordinates are -2 or 2 (and none whose y-coordinate is -6). The x-coordinates of these points are not integers, but we can use the symmetry of the graph (with respect to the vertical line x=-2) to deduce that if these points are (x1,-2), (x1,-2), (x3,2), and (x4,2), then x1+x2=-4 and x3+x4=-4. Therefore, the sum of all four x-coordinates is -8.

For goodness sakes Chris, stop putting full answers after people have tried to teach something.

You are preventing children from learning and you are facilitating cheating!

Then, you are left with 18 boys with uncovered left hands, so they must be paired to hold the right hands of the girls.

Well, the answer is 170,970,456,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000.

Just move the decimal 308 places.

LOL Rom.

With my warped sense of humour you have me in stitches.