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im so confused

We can find DZ by using the formula for the area of a trapezoid

120 = (1/2) (DC) (WC + DZ)      and      (DC = AB)       so we have

120  = (1/2) (12) (6 + DZ)

120 =  6 (6 + DZ)

20 =  6 + DZ

14  = DZ = BW

Note that, by ASA,  triangle BQW is cogruent to triangle DQZ

So....their altitudes are equal  =  (1/2)AB   =  6

So.....the area of triangle BQW   =  (1/2)BW (6)  = (1/2)(14)(6) = (1/2)84 =  42 units^2

This is it

0. Any number times 0 is 0 and 0 rational.

What are you looking for?

What was the answer????????

Thank you!!! :)

d/dx (sin2x) = 2 cos 2x

can't you do ANY of your own thinking?

Sure, it's possible by considering the area. We can assume that the legs have a length 1 and so we wish to find the length of the hypothenuse which we will call x. The area then becomes $\frac{1}{2}$. If we draw the height of the triangle from the hypothenuse we notice that it has the length $\frac{x}{2}$ and so the area can also be calculated to be $\frac{x \cdot \frac{x}{2}}{2}$. We then get the equation $\frac{x^2}{4}=\frac{1}{2}$ or $x^2=2$ and so $x= \sqrt{2}$ and the ratio is $1:1:\sqrt{2}$

so c?

V of sphere = 4/3 pi r^3

4/3 pi r^3 = 36 pi      divide both sides of eq by pi

4/3 r^3 = 36              multiply by 3/4

r^3 = 27           Cube root

r = 3

diameter = 2*r = 6 cm

5000 * .15  x 1/2 = $375 each 6 months

Tan = sin/cos

1/tan = cos/sin = cot

CPhill, you and Melody really are the best, you guys are moderators!

They are currency for a very popular American game called Fortnite

Yeah, I know that one, it's cool!

Figure below? I don't see a figure...

Very odd question but okay...

289585986546546757865978689576+25899967379693475634796573647563534657346576678768967 = 

25,899,967,379,673,475,634,796,863,233,550,081,204,104,442,657,458,543.

Wow, that was a very odd question!

That's precisely

Twenty-five sexdecillion, eight hundred ninety-nine quindecillion, nine hundred sixty-seven quattuordecillion, three hundred seventy-nine trecillion, six hundred seventy-three dodecillion, four hundred seventy-five undecillion, six hundred thirty-four decillion, seven hundred ninety-six nonillion, eight hundred sixty-three octillion, two hundred thirty-three septillion, five hundred fifty sextillion, eighty-one quintillion, two hundred four quadrillion, one hundred four trillion, four hundred forty-two billion, six hundred fifty-seven million, four hundred fifty-eight thousand five hundred forty-three.

WOW!

You are very welcome!

:P

Yours is a much simpler approach Rom, and, yes, I totally forgot about the "integer coefficients" requirement.

​ Method of difference

Formula:

$\begin{array}{|rcll|} \hline \displaystyle \sum \limits_{r=1}^{n}r^2 &=& \dfrac16n(n+1)(2n+1) \\\\ \displaystyle \sum \limits_{r=1}^{2n}r^3 &=& \dfrac1{2^2} (2n)^2(1+2n)^2 =n^2(1+2n)^2 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline &&\mathbf{ \displaystyle \sum \limits_{r=1}^{n}(6r-3)^2 } \\\\ &=& \displaystyle \sum \limits_{r=1}^{n}(36r^2-36r+9) \\\\ &=& \displaystyle 36\sum \limits_{r=1}^{n}r^2- 36 \sum \limits_{r=1}^{n}r+ 9\sum \limits_{r=1}^{n}1 \\\\ &=& 36\dfrac16n(n+1)(2n+1) -36\dfrac{(n+1)}{2}n +9n \\\\ &=& 6n(n+1)(2n+1) -18n(n+1) +9n \\\\ &=& 6n(n+1)(2n+1) -18n^2-18n+9n \\\\ &=& 6n(n+1)(2n+1) -18n^2-9n \\\\ &=& 3n\Big( 2(n+1)(2n+1) -6n-3 \Big) \\\\ &=& 3n\Big( 2(n+1)(2n+1) -3(2n+1) \Big) \\\\ &=& 3n(2n+1)\Big( 2(n+1) -3 \Big) \\\\ &=& 3n(2n+1)(2n-1) \\\\ &\mathbf{=}& \mathbf{3n(4n^2-1)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{\displaystyle \sum \limits_{r=1}^{2n}r^3 - \sum \limits_{r=1}^{n}(6r-3)^2 } \\\\ &=& n^2(1+2n)^2 - 3n(4n^2-1) \\ &=& n^2(1+2n)^2 - 3n(2n-1)(2n+1) \\ &=& n(1+2n)\Big(n(1+2n) - 3(2n-1)\Big) \\ &=& n(1+2n)(n+2n^2 - 6n +3 ) \\ &=& n(1+2n)(2n^2 -5n +3 ) \\ &\mathbf{=}& \mathbf{ n(1+2n)(2n-3)(n-1) } \\ \hline \end{array}$

or you can just find

 $p\left(\dfrac 1 x\right) = 0\\ \left(\dfrac 1 x\right)^3 - \dfrac 2 x - 3 = 0\\ 1 - 2x^2 - 3x^3 = 0 \\ \text{This is the form the problem asks for, but}\\ \text{dividing both sides by -3}\\ x^3 + \dfrac 2 3 x^2 - \dfrac 1 3 = 0\\ \text{which is the form Alan wrote it in}$

$\text{similarly for the second part}\\ x=\dfrac{1}{2\alpha+1} \Rightarrow \alpha = \dfrac 1 2\left(\dfrac 1 x -1\right)\\ \text{so find }\\ p\left(\dfrac 1 2\left(\dfrac 1 x -1\right)\right)\\ \text{and work the same sort of algebra to get nice coefficients}$

On the assumption that a) is indeed  $\frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma}$ we have: