All Answers

+118723

y^3/8

thats the answer

MelodyAug 28, 2014

+118723

+10

$$Let the triangle have side length 2a, the perpendicular height will be $a\sqrt3$\\
Let the depth of prism be D (all units in cm)\\
A $1cm^3$ container has the capacity of $1mL$\\
Volume of prism = Area of triangle $\times$ Depth\\
$5178 = \frac{1}{2} \times 2a \times a\sqrt3 \times D$\\
$5178 = a^2\sqrt3 \times D$\\\\
$D=\dfrac{5178}{a^2\sqrt3} $\\$$

$$$D=\dfrac{5178}{a^2\sqrt3} $\\\\
Surface area\\
S=2(triangles)+3(rectangles)\\\\
$S=2(\frac{1}{2}\times 2a \times a\sqrt3)+3(2a\times D)$\\\\
$S=2(a^2\sqrt3)+3(2a\times \frac{5178}{a^2\sqrt3})$\\\\
$S=2\sqrt3a^2+\frac{5178*6}{a\sqrt3}$\\\\
$S=2\sqrt3a^2+\frac{31068a^{-1}}{\sqrt3}$\\\\$$

$$Surface area will be minimum when \\\\
$\frac{dS}{da}=0\;\;and\;\;\frac{d^2S}{da^2}>0$\\\\
$\frac{dS}{da}=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\
$0=4\sqrt3a-\frac{31068a^{-2}}{\sqrt3}$\\\\
$0=12a-31068a^{-2}$\\\\
$0=12a^3-31068$\\\\
$12a^3=31068$\\\\
$a^3=2589$\\\\
$a\approx 13.73$cm\\\\\\
$\frac{d^2S}{da^2}=4\sqrt3+\frac{2*31068a^{-3}}{\sqrt3}>0\;for\; all \;a>0$\\\\
Therefore any turning point must be a minimum.$$

$$\\a\approx 13.73cm\\\\
D\approx \frac{5178}{13.73^2*\sqrt3}\\\\
D\approx 15.86cm\\\\\\
\mbox{The triangle has side length 2*13.73= 27.46cm and the prism is 15.86cm deep}\\\\
check\\
V=1/2*27.46*13.73*\sqrt3*15.86 = 5178.5cm^3 \qquad good$$

I think that is okay

MelodyAug 28, 2014

i have now ideA

GuestAug 28, 2014

What I meant was, you can only use three once and four twice...

GuestAug 28, 2014

+11912

Your Welcome Melody!

And................im happy you liked my post Aziz!thank you!

rosalaAug 28, 2014

+118723

(10/2)^2 is the number that must be tacked on.

x^2 +10x +25

MelodyAug 28, 2014

+118723

inverse sine is the same as arcsine

just use asin

MelodyAug 28, 2014

+118723

(3+4)*(3+4)*(3-4/4)+(3-4/4)=100

MelodyAug 28, 2014

+118723

The technique is the same as the one I just did.

Look here and try and figure it out for yourself.

http://web2.0calc.com/questions/quadratic-functions-derived-from-vertex-and-a-point#r1

MelodyAug 28, 2014

+118723

I just answered this

http://web2.0calc.com/questions/quadratic-functions-derived-from-vertex-and-a-point

MelodyAug 28, 2014

Vieta's formulas are formulas that relate the coefficients of a polynomial to sums and products of its roots ...

http://en.wikipedia.org/wiki/Vieta's_formulas

Heureka

Very cool Heureka!

Heureka

heureka

GuestAug 28, 2014

+26400

+10

Hi Melody,

François Viète (Latin: Franciscus Vieta; 1540 – 23 February 1603), Seigneur de la Bigotière, was a French mathematician whose work on new algebra was an important step towards modern algebra, due to its innovative use of letters as parameters in equations

The Laws

Basic formulas

Any general polynomial of degree n

$P(x)=a_nx^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0 \,$

(with the coefficients being real or complex numbers and a_n ≠ 0) is known by the fundamental theorem of algebra to have n (not necessarily distinct) complex roots x₁, x₂, ..., x_n. Vieta's formulas relate the polynomial's coefficients { a_k } to signed sums and products of its roots { x_i } as follows:

$\begin{cases} x_1 + x_2 + \dots + x_{n-1} + x_n = -\tfrac{a_{n-1}}{a_n} \\ (x_1 x_2 + x_1 x_3+\cdots + x_1x_n) + (x_2x_3+x_2x_4+\cdots + x_2x_n)+\cdots + x_{n-1}x_n = \frac{a_{n-2}}{a_n} \\ {} \quad \vdots \\ x_1 x_2 \dots x_n = (-1)^n \tfrac{a_0}{a_n}. \end{cases}$

Equivalently stated, the (n − k)th coefficient a_n−k is related to a signed sum of all possible subproducts of roots, taken k-at-a-time:

$\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} x_{i_1}x_{i_2}\cdots x_{i_k}=(-1)^k\frac{a_{n-k}}{a_n}$

for k = 1, 2, ..., n (where we wrote the indices i_k in increasing order to ensure each subproduct of roots is used exactly once).

The left hand sides of Vieta's formulas are the elementary symmetric functions of the roots.

Example

Vieta's formulas applied to quadratic and cubic polynomial:

For the second degree polynomial (quadratic) $P(x)=ax^2 + bx + c$ , roots $x_1, x_2$ of the equation $P(x)=0$ satisfy

$x_1 + x_2 = - \frac{b}{a}, \quad x_1 x_2 = \frac{c}{a}.$

The first of these equations can be used to find the minimum (or maximum) of P. See second order polynomial.

For the cubic polynomial $P(x)=ax^3 + bx^2 + cx + d$ , roots $x_1, x_2, x_3$ of the equation $P(x)=0$ satisfy

$x_1 + x_2 + x_3 = - \frac{b}{a}, \quad x_1 x_2 + x_1 x_3 + x_2 x_3 = \frac{c}{a}, \quad x_1 x_2 x_3 = - \frac{d}{a}.$

heurekaAug 28, 2014

+118723

The second question is really puzzling me and I would like more input from other mathematicians.

I have made this graph and it appears to be telling me that one approx answer is h=4.2 and k=-1.6

but it is confusing. Would anyone like to comment.

https://www.desmos.com/calculator/0wieewtzra

MelodyAug 28, 2014

#14

+1832

Thanks for all ... alan and melody

xvxvxvAug 28, 2014

+118723

$$\\Pa=\dfrac{w(v-t)}{g}\\\\
gPa=wv-wt\\\\
gPa-wv=-wt\\\\$$

And you can finish it

MelodyAug 28, 2014

+118723

$$\\\frac{16}{a}=\frac{2}{3}\\\\
\frac{a}{16}=\frac{3}{2}\\\\
a=\frac{3\times 16}{2}\\\\
a=24\\\\$$

MelodyAug 28, 2014

+118723

Thank you Heureka

What is Vieta

MelodyAug 28, 2014

#98

+118723

Fri 29/8/14

1) Ninja presents his full answers beautifully - just look at this equation solution !!

http://web2.0calc.com/questions/8-x-1-4x-8

* 2) Today Aziz suggested this "Learning Calculus" website. People should never forget that there are many really good information sites available on the web. Thanks Aziz.

https://www.khanacademy.org/math/differential-calculus

3) More questions here! (I included this last time too but it has been added to.)

http://web2.0calc.com/questions/quadratic-functions-derived-from-vertex-and-a-point

4) Minimizing question. We don't get many of these.

http://web2.0calc.com/questions/working-out-the-triangle-length-and-prism-length-of-a-equilateral-triangular-prism#rr1

* 5) Great Simultaneous equations - elimination method solution by Ninja

http://web2.0calc.com/questions/solve-2x-3y-7-x-y-3

* 6) A little Physics anyone? Thanks Alan

http://web2.0calc.com/questions/physics_65

7) Unanswered puzzle

http://web2.0calc.com/questions/how-do-you-get-100-with-only-the-numbers-3-4-and-4

Using the Desmos calculator to graph inequalities. (thanks Chris)

http://web2.0calc.com/questions/graph-the-inequality-2-y-2-2-y-4

8) Rosala's laugh. Thanks Rosala

http://web2.0calc.com/questions/let-s-end-the-day-with-a-laugh_1

9) Good fluid solution question (thanks Chris and Alan)

http://web2.0calc.com/questions/how-much-pure-acid-should-be-mixed-with-3-gallons-of-40-acid-solution-in-order-to-get-a-90-acid-solution

10) I haven't had a chance to look at this yet but it might be interesting.

http://web2.0calc.com/questions/vedic-maths-worth-looking-at

11) This advanced maths has been included before but Heureka has added to it. (Thanks Heureka)

http://web2.0calc.com/questions/trigonometry_78

♬ ♬ ♬ MELODY ♬ ♬ ♫♪ You light up my life ♪ ♫

MelodyAug 28, 2014

+33661

First rearrange by adding y to both sides:

2x = 1 + y

Divide both sides by 2:

x = (1 + y)/2

Now substitute 7 for y:

x = (1 + 7)/2

or x = 4

AlanAug 28, 2014

#268

+118723

@@ End of Day Wrap : Thurs 28/8/14 Sydney, Australia Time 9:55pm ♬

Good evening, afternoon, morning all,

Our intrepid answerers today were Will8537, CPhill, NinjaDevo, TherealQuestioner, AzizHusain, Rosala, TakahiroMaeda, BluePhoenix912, Alan, Kitty<3, DragonSlayer554 and GoldenLeaf. Thank you all.

Yesterday I requested that people think before they give full answers as the aim is to teach not to do people's homework for them. I saw a number of answerers respond very positively to this request. Thank you, I am really pleased, I am sure that we all want our answers to promote maximum learnning.

The sticky threads "Great answers to Learn from" and "Reference Material" were beautifully put together and kept updated by NinjaDevo. This has been a huge benefit to me and to everyone who uses the forum.

Unfortunately for us Ninja has now resigned from this job. School will be starting again for him next week and he will be to busy for this extra responsibility. I want you to know Ninja that I am extremely grateful for all that you have done and I seriously hope that you continue to have a very strong presence in the forum.

If anyone would like to volunteer to help me with tasks like these I would be very grateful.

Now for some interest threads:

1) An interesting function query.

http://web2.0calc.com/questions/pre-calc-functions

2) Rosala's laugh

http://web2.0calc.com/questions/let-s-end-the-day-with-a-laugh

3) A good one for the younger crowd

http://web2.0calc.com/questions/for-all-possible-values-of-a-if-a-b-2-a-2-what-must-b-equal

4) I like doing things the long way lol

http://web2.0calc.com/questions/quadratic-functions-derived-from-vertex-and-a-point

That is it for my day. Enjoy the rest of yours.

♫♪ ♪ ♫ ♬ ♬ MELODY ♬ ♬ ♫♪ ♪ ♫

MelodyAug 28, 2014

+33661

Yes. Click the 2nd button and you will see mod appear.

AlanAug 28, 2014

+118723

This is a parabola, there is an infinite number of x,y coordinates that will make this true.

Perhaps you actually want the solutions to

0=x^2-x-6

0=(x-3)(x+2)

x=3 or x=-2

The roots of y=x^2-x-6

are (3,0) and (-2,0)

MelodyAug 28, 2014

+118723

2) This second one looks tougher.

I assume you would use simultaneous equations. Lets give that a go.

containing (0,6) and (3,-1)

$$\\y=a(x-h)^2+k\\\\
6=a(0-h)^2+k\\
6-k=ah^2 \qquad (eqn\;1)\\\\
-1=a(3-h)^2+k\\
-1=a(9+h^2-6h)+k\\
-1-k=ah^2-6ah+9a \qquad(eqn\:2)\\\\
6-k=ah^2 \qquad (eqn\;1)\\
-1-k=ah^2-6ah+9a \qquad(eqn\:2)\\
eqn1-eqn2\\
6-k-(-1-k)=ah^2-(ah^2-6ah+9a)\\
7=6ah-9a\\
7=a(6h-9)\\
a=\dfrac{7}{6h-9}\\\\$$

I've been writing this and putting it into LaTex at the same time - there could easily be errors.

anyway. I am assuming that from here you can go back sub this value of a into the simultaneous equations and come up with a value for h and k.

Why don't you give it a go and see if you can finish it? I'll look at it some more if you get stuck.

MelodyAug 28, 2014

its gonna be 5.250km + 3.750km + 4.050km = 13.050km (or 13050 meters or 13Km and 50 km)

GuestAug 28, 2014

All Answers 358108 Answers

The Laws

Basic formulas

Example

0 Online Users

Top Users

Sticky Topics