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I have figured it out! Thank you so much for your help.

- Daisy

See the answer here:  https://web2.0calc.com/questions/question_139

See if you can work it out from that. Let us know either way :)

I believe the area if 28. 

I am sorry, right now I don't have time to write it up (and hence check my own logic) 

but

I used similar triangles. 

If you let M be the midpoint of AC then   triangle AMB= triangle CMB

and they are both similar to triangle ACD

ACD is bigger by a factor of 2.

hence its area is bigger by a factor of 4

So the big one has twice the area of the 2 little ones combined

so

Area of ACD = 2/3 * 42 = 28 unites squared.

Haha! Thanks for informing me, Melody! I will keep that in mind in the future.

- Daisy

In order to solve an equation like this, we need to isolate the variable. To do this, we need to get rid of \(5\) from \(x\). Since \(5\) is being multiplied to \(x\), we need to do the opposite operation, which is division. When we divide \(5\) from \(x\), we also have to divide \(5\) from \(25\). When \(5\) divides \(25\)​, we get \(5\), so your result is \(x = 5\).

- Daisy

I'm not sure....but here's my best effort

Let ABCD  be the  trapezoidal field

Its  area   = 

(1/2)(height) (DC + AB) = 

(1/2)(50√3) (100 + 200) =

7500√3   m^2  

And  the  crop will be brought to the longer side whenever it is contained within  the trapezoid  AEFB

Reasoning :

DB bisects  angle CBA....angle FIB  = FKB  and FB  = FB

So....by AAS, triangle FIB  = triangle FKB

So....any crop harvested within triangle FIB will be closer to IB than BK

For the same reason any crop harvested within triangle EJA  will be closer to JA  than side AL

And any crop harvested within EFIJ  will be closer to the longest side than to side DC

And the area of AEFB  =

(1/2)(EJ) (AB + EF)  =

(1/2)(25√3) (200 + 50) =

3125√3  m^2

So  the fraction of the crop brought to the longest side  is

3125√3  / 7500√3  =   5/12

Guys I could really use some help here!!

To compute: \(\displaystyle\int^{\infty}_{0}\dfrac{\ln^2\left(z\right)dz}{z^4+z^2+1}\).

I think this does come out by Cauchy, but it's quite heavy.

First you need the poles created by the zeros of \(z^4+z^2+1=0\): first put \(Z=z^2\) so \(Z^2+Z+1=0.\) The roots of this equation are the complex cube roots of unity \(\omega_1={1\over2}+i\sqrt{3\over2}\) and \(\omega_2=\overline\omega_1=\omega_1^2\).

Therefore the roots of \(z^4+z^2+1=0\) are the square roots of \(\omega_1,\ \omega_2\). But since \(\omega_2^2=\omega_1\)and \(\omega_1^2=\omega_2\) these four square roots are just \(z=\pm\omega_2\) and \(z=\pm\omega_1\).

You then have to compute the residues at these 4 poles: it helps that \(\omega_1=\exp(2\pi i/3)\) being a cube root of unity has absolute value 1 which simplifies the computation of \(\ln(z)\).

Having got the residues you then do the usual contour integration trick round two circles radius R and r joined by a cut up the negative real axis. This very much follows the example given in Wikipedia for the case where the denominator is just \((1+z^2)^2\) given here:

Note: this is not an answer.

Hi Daisy,

This shape is ony called a Trapezoid in USA and Canada. The rest of the English speaking world knows it as a Trapezium.

In England a Trapezoid has no parallel sides. 

To me, in Australia, a trapezoid has no meaning at all.   At least I have never known of one.

Anyway, you are referring to a quadrilateral with one pair of parallel sides.    (What I would call a Trapezium)

Why would the horizontal asymptote mean that it it set to infinity and why would it be equal to a and -4? Could you please explain? Sorry, I really don’t understand...,.

You can first find the second \(a\), by setting up an equation and solving it. We can make the equation \(\frac{2}{5} = \frac{a}{3}\), which we can solve using cross-multiplication. Then, we get \(6 = 5a\), which gives us \(a = \frac{6}{5}\). So if \(a = \frac{6}{5}\), and \(a\) is inversely proportional to \(c\), than we can see that \(\frac{6}{5} \times c\) must equal \(18\), because \(2 \times 9 = 18\). \(\frac{18}{\frac{6}{5}} = 18 \times \frac{5}{6} = 15\). So \(c = 15\).

- Daisy

Nevermind, I have figured it out!

- Daisy

If the   quadratic function opens downward, we will have a negative sign in front of the x^2  term...so we can eliminate the last two answers right away

In the form     ax^2  + bx + c, the  x coordinate of the vertex  =  -b / [ 2a]

Look at answer B.....letting a  = -1   and b  = 4....the x coordinate of the vertex  is

-4  / [ 2 * -1 ]   =  -4 / -2   =   2

To find the y value associated with this x value, put this "2"  back into the function

-(2)^2  + 4(2) +   1   =

-4  +   8  +   1    =

-4  +  9   =

5

So....this function  turns downward  and has a vertex  of  (2, 5)

Here's a graph to confirm this  :  https://www.desmos.com/calculator/tlavaymezd

yup i got it a long time ago i forgot the answer

r(x)  =  1 / (1 - x)

r(30)  =  1 / ( 1 - 30)  = -1/29

r(r(30))  = r(-1/29)  =    1 / (  1 -  -1/29)  = 1 / (1/ 30)  = 30

r(r(r(30))) =  r(30)  =  1 / (1 - 30)  =  -1/29

Note  that the pattern repeats....so when r is applied 6 times we will end up with  -1/29

Hi Max: Here is an online "Integral Calculator" that appears to be able to solve your Integral with many steps as well: https://www.integral-calculator.com/

Good luck.

Thanks!

Thanks!

"A function f has a horizontal asymptote of y = -4 a vertical asymptote of x = 3 and an x-intercept at (1,0). Part (a): Let f be of the form f(x) = \frac{ax+b}{x+c}.Find an expression for f(x). "

\(f(x)=\frac{ax+b}{x+c}\)

Horizontal asymptote means set x to infinity so \(f(x)\rightarrow a\)  hence we immediately have \(a=-4\)

Vertical asymptote means set denominator to zero at x = 3, so \(3+c=0 \text{ or } c = -3\)  

x intercept at (1, 0) means \(\frac{a\times1+b}{1+c}=0\)  so  \(\frac{-4+b}{1-3}=0 \text{ or }b = 4\)  

Hence \(f(x)=\frac{-4x+4}{x-3}\text{ or }f(x)=\frac{4(1-x)}{x-3}\) 

Use the same approach for part b.

(Edited to correct silly mistake!  Thanks heureka.)

THANK YOU SOOOO MUCH!!!!!!!

Thanks. It’s a method I’ve only seen used on cubic equations before - to eliminate the x² term. I guess the quartic specified here was deliberately chosen to simplify enough that it would be easy to find the roots of the resulting transformed function.

Determine a constant k such that the polynomial

\(\displaystyle P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2)\)

is divisible by x+y+z

\(\begin{array}{|clcll|} \hline \\ P(x, y, z)\text{ is divisible by } x+y+z \\ \\ \hline \\ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = (x+y+z) \cdot f(x,y,z) \\ \text{Set } x+y+z = 0 \\ \text{so:} \\ P(x, y, z) = x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = 0 \cdot f(x,y,z) \\ x^5 + y^5 + z^5 + k(x^3+y^3+z^3)(x^2+y^2+z^2) = 0 \\ \\ \text{Set arbitrary } x+y+z = 2-1-1= 0 \qquad \text{so } x=2, \quad y=-1, \text{ and } z=-1 \\\\ P(2, -1, -1) = 2^5 + (-1)^5 + (-1)^5 + k\left(2^3+(-1)^3+(-1)^3\right)\left(2^2+(-1)^2+(-1)^2\right) = 0 \\ 32-1-1 + k(8-1-1)(4+1+1) = 0 \\ 30 + k\cdot 6 \cdot 6 = 0 \\ 30 +36k = 0 \\ 36k = -30 \\ k = -\dfrac{30}{36} \\ \mathbf{ k = -\dfrac{5}{6} } \\ \hline \end{array}\)

Example:

\(\begin{array}{|rcll|} \hline x&=& 1 \\ y&=& 2 \\ z &=& 3 \\ x+y+z &=& 6 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline P(1, 2, 3) &=& 1^5 + 2^5 + 3^5 - \dfrac{5}{6}\left(1^3+2^3+3^3\right)\left(1^2+2^2+3^2\right) \\ &=& 1 + 32+ 243 - \dfrac{5}{6}\cdot(1 +8+27)(1+4+9) \\ &=& 276 - \dfrac{5}{6}\cdot 36\cdot 14 \\ &=& 276 - 5\cdot 6\cdot 14 \\ &=& 276 - 420 \\ &=& -144 \\ &=& -24\cdot 6 \quad \text{This is divisible by } x+y+z = 6 \\ \hline \end{array}\)

yes can you please solve it?

Thanks, i too am still confused.