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Suppose that x,y,z are positive integers satisfying x is less than or equal to y is less than or equal to z, and such that the product of all three numbers is twice their sum. What is the sum of all possible values of z?

I just made a possible list and then cancelled the ones that didn't work.   I was left with 3 possible triads

You know    \(x\le y\le z\qquad \text{Where that are all positive integers}\)

and you know

\(xyz=2(x+y+z)\)

the smallest the RHS can be is 6 and the biggest is 54

So the product of x, y and z is between 6 and 54

Now I just listed all the possibilities for the triads that multiply to between 6 and 54 and then cancelled them out if they did not fit the equality.

Give it a go.

the smallest one is

1,1,6

then

1,1,7

etc.

Try doing it yourself.

One factor of this polynomial is (2x − 1). 2x^3+5x^2-27x+12. Use long division to determine which expression represents the other factor.

Hello Guest!

\((2x^3+5x^2-27x+12)\) : \((2x-1)\) = \( x^2+3x-15\) 

\(\underline{2x^3-\ x^2}\)

            \(6x^2-\ 27x\)

            \(\underline{6x^2\ +\ 3x}\)

                     \(-30x +\ 12\)

                     \(\underline{-30x+15}\)

                                    \(-3\)

\((2x^3+5x^2-27x+12)\) = \( (x^2+3x-15\) - \(\large \frac{3}{2x-1}\)) \(\times (2x-1)\) 

  !

Do you mean:

\(f(x)=2x^2−5x−123x^2−11x−4\\ f(x)=−121x^2−16x−4\\\)

This is a parabola, there are no holes.

so we have inverse is x=4y+16,so x-16=4y, so 1/4x-4=y. the functions are inverses.

What is the area of this figure?

Hello Guest!

\(A=9\times 3\ dm^2+(9+7)\times 4\ dm^2+(9+7-4-2)^2\times\frac{3.14}{4\cdot 2}\ dm^2\)

\(A=130.25 \ dm^2\)

  !

Area = 7*9 + pi*(5^2/2) + 7*4 = 91 + 25/2*pi

the idea is spliting your figure to 2 rectangles and 1/2 circle.

(A,B) = (7,4)

It's me again.  I made a typo.  The third line should read  

                                               the last number on page 3 would be 751 ~ aka 750 + 1   

The series starts with 1 and then counts up by 5's.  

There are 50 numbers to a page so the last number on page 1 would be 251 ~ aka 250 + 1 

                    add the next 50 x 5 and the last number on page 2 would be 501 ~ aka 500 + 1  

                                                          the last number on page 3 would be 551 ~ aka 750 + 1

                                                          and so on.  

We can see that the last number on any page is (the page number times 250) + 1  

So, which page would contain 12346 

If you divide 12346 by 250 you get 49.384 so that locates 12346 on the first page past 49, or page 50  

_.

If f(x) = 3x^2 - 5, what is the value of f(f(-1))?

Hello Guest!

\(f(x) = 3x^2 - 5\\ f(-1)=3\cdot (-1)^2-5 \)

\(f(-1)=-2\)

\(f(f(-1))=f(-2)=3\cdot (-2)^2-5\)

\(f(f(-1))=7\)

  !

ABCD is a quadrilateral such that AB\(\parallel\)CD. We know that angleA= angle B= 60 degrees and AD=CB=1 . If AB=10 , what is DC?

Hello Guest!

\(\overline{DC}=\overline{AB}-2\cdot (1\cdot cos\ 60^{\circ})=10-2\cdot 0.5\)

\(\overline{DC}=9\)

  !

top portion   3 x 4 x 10

bottom portion  3 x 4 x 4     cm³         Add together

The peak of the wave is usually at  zero...   it is shifted to the right to/by  pi/4

In short, the answer is 143cm².

This is how I did it: 

First, we must find the surface area of every face.

10x5 = 50

6x8 = 48 (we can do this because both triangles are the exact same)

5x8 = 40

6x5 = 30

Now we add all this up and get 143cm².

Hope it helps!

What are the coordinates of the hole in the function?

f(x)=2x2−5x−123x2−11x−4

Is anyone able to help me with this? Would greatly appreciate it!

 A(0, 3)      B(12, 0)       origin O       center of a circle  C   

1/     Line segment  AB = sqrt(AO² + BO²)

 2/    Angle (ABO) = arctan(AO / BO)              ∠ABO = ∠BAC

3/     Radius  AC = (AB/2) / cos∠BAC

a = # apples

r = # rasberries = a +32      summed, they equal 108

2.25 a    +   1.50 ( a+32)    =   108      solve for a       then   r = a + 32

Finding trigonometric expression

sec (arctan x/2) 

Hello  Lolaaaaaa!

\(f(x)=sec (arctan\ \frac{x}{2})=\frac{1}{cos\ (arctan\ \frac{x}{2})}\)

   x :  1        2        3          4        5         6         7         10 

f(x) : 1.118 1.414  1.803  2.236  2.693  3.162  3.640  5.100 

  !

Area =  1/ 2 base * height

Base goes from  -4  to  4    a distance of 8

Height goes from -3 to + 3  a distance of 6

h*g =  (4x+1)(-4x+1)  = -16x^2 +1     now put in   x/3   for x

                                   -16x^2/9 +1    and put in a common denominator of 9

                                   (-16x^2 +9 ) / 9    and re-arrange

                                      (9 - 16x² ) / 9

You are given the law and all of the variables....just plug in and compute

9 x10^9  *  4.2 x a0^-9  *  1.10 x 10 ^-9  / (.005)^2 = ........ Newtons