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Note that QCA is a right triangle....area = 1/2 *  base * height

1/2 base * height = area

1/2 * 3 * (12-p) = area

18 - 3/2 p = area

When the ball hits the ground, height = y   will be ZERO

0 = - 16t^2 - 80t + 54         Use quadratic equation    a = -16      b = -80    and   c = 54   to solve for t

                                                     throe out the negative 't' value

\(t = {80 \pm \sqrt{(-80)^2-4(-16)(54)} \over 2(-16)}\)

You need to know how to calculate using the Quadratic Formula......if you already do, then here is a quick and easy Quadratic Formula calculator that I often use :

https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php

-2(s + 7) > 4s + 8      expand L side

-2s-14 > 4x+8            add 2s and 12 to both sides of the equation

0 > 6x+ 22                 subtract 22 from both sides

-22 > 6x                     divide both sides by 6

-22/6 >x                     or...re-wrote as

x< - 22/6                      (-inf, - 22/6)

Let price of fan = rs X

Price of table lamp =rs (2/3) X

Total money spent on 12 table lamps & 2 fans

= 8X + 2X

= 10X

Let total money with him be Y

(3/5)Y = 10X

Y= 50X/3

Cost of table lamp = (2/3) X

If he had spent all his money on table lamp

Total table lamp = 25

For second part

Money left , (2y/5) = (20X/3)

Money spent on 4 fans = 4X

Money left = 8X/3

With this total table lamp = 4

So he bought = 12+4 = 16 table lamp altogether

Therefore total table lamp he bought altogether = 16

Good spotting EP   

Are you on AoPS? If so I would recommend asking there. I'm OlympusHero and you can get answered fast to this harder stuff there, though there are some really talented people here (melody, cphill, etc... I'm looking at you :P)

Please, what does it say on Vieta, heureka?

Let Carl had X marbles

Marbles with Alan = X-24

Let marbles with ben = Y

Marbles with Alan = (X+Y)/3

X-24 = (X+Y)/3

3X-72 = X+Y

2X-Y= 72...(1)

Marbles with Ben =(2X-24)/5

5Y= 2X-24

2X-5Y= 24...(2)

Solving (1) & (2)

Y= 12

X= 42

Total marbles with Alan = 18

Total marbles with all of them = 12+18+42

= 72

deez nuts

This is not possible in a right triangle....   the shorter side + the 20 side do not add up to more than 99 ....necessary to meet the triangle side length rule. .

Normally, you cannot have the sqrt of a negative number....so the domain  is

[0,+inf)

Well, I drew the pic, and it does, on my graph anyway  (I replaced P with A' )

The roots of \(7x^2 + x - 5 = 0\) are \(a\) and \(b\).
Compute \(\dfrac{1}{a} + \dfrac{1}{b}\).

By Vieta:

\(\begin{array}{|rcll|} \hline 7x^2 + x - 5 &=& 0 \quad | \quad : 7 \\ x^2 + \underbrace{\frac{1}{7}}_{=-(a+b)}x \underbrace{-\frac{5}{7}}_{=ab} &=& 0 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline a+b &=& -\dfrac{1}{7} \\\\ ab &=& -\dfrac{5}{7} \\\\ \hline \dfrac{a+b}{ab} &=& -\dfrac{1}{7} \over -\dfrac{5}{7} \\\\ \dfrac{a}{ab}+\dfrac{b}{ab} &=& \dfrac{1}{7} \over \dfrac{5}{7} \\\\ \dfrac{1}{a}+\dfrac{1}{b} &=& \dfrac{1}{7} \times \dfrac{7}{5} \\\\ \mathbf{\dfrac{1}{a}+\dfrac{1}{b}} &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}\)

The roots of 7x^2 + x - 5 = 0 are a and b. Compute 1/a + 1/b.

Hello Guest!

\(7x^2 + x - 5 = 0\\ x^2+\frac{1}{7}x-\frac{5}{7}=0\\ x=-\frac{1}{14}\pm \sqrt{\frac{1}{196}+\frac{5}{7}}\\ x=-\frac{1}{14}\pm\sqrt{\frac{1+5\cdot 28}{196}}\\\)

\(x=-\frac{1}{14}\pm \frac{1}{14}\sqrt{141}\\ a=-\frac{1+\sqrt{3\cdot 47}}{14}\\ b=-\frac{1-\sqrt{3\cdot 47}}{14}\\ \)

\(\frac{1}{a}+\frac{1}{b}=-\frac{14}{1+\sqrt{3\cdot 47} }-\frac{14}{1-\sqrt{3\cdot 47} }\\ =\frac{-14\cdot (1-\sqrt{3\cdot 47}+1+\sqrt{3\cdot 47})}{1-3\cdot 47}=\frac{-28}{-140}\)

\(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{5}\)

  !

Also answered here

just use Pythagoras's theorem to work it out.  The Pythagorean triple that they gave you is irrelevant.

Hint:

Since you want the shortest side, there will be a take away.  (minus)

Recall that a perfect square is the square of some integer. How many perfect squares less than 1,000 can be represented as the difference of two consecutive perfect squares?

so... how many Pythagorean triads are there such that the smallest one squared is less than 1000

Pythagorean triads with a prime number

sqrt(1000) = 31.6227766016837933

I get 16 but there is probably a much better way to do it.

 Let ABC be a triangle with AB = 13, BC = 14, and AC = 15. The incircle

of ABC touches AB and AC at points D and E, respectively. The triangle’s
A−excircle (tangent to segment BC and the extended rays AB and AC, outside of
the triangle) touches BC at P. Find [ADE]/[BDP].

AND

I need to see how you attempt my teaching questions that I have asked you.

It is no wonder you cannot learn if you do not even attempt to follow your teacher's (my) instructions.

That is very nice interaction GLAZ3D.

Welcome to 'our'   web2.0calc forum   

Hi Why,

In case you did not realize, you are allowed/encouraged to give yourself a point for a good answer.

AX = sqrt(AC² + CX²)

That was easy.

What's the length of CY?

Answer: \(26.9230769231\)

Solution:

Sorry for the late response. It is somewhere between 26 and 27, because \(\sqrt700\) is somewhere between 26 and 27. Which means the floor of x is 26. So 26x=700. This turns into x = 700/26. Doing the division gives \(26.9230769231\).

Yes, but if a person doesn't put any effort into presenting their question properly, do you think they being polite, or that they are really interested in learning anything?

I do not want to do homework for people who are not polite and are not attempting to learn.

Answer: \(1.5\sqrt2\)

Solution:

If BC and DC are equal to each other (they are both three), then ABCD must be a square. BD would just be a diagonal of said square, splitting it in half. CE at that ponit is obviously just 1/2 of BD, and BD is \(3\sqrt2\) (\(\sqrt(3\cdot3+3\cdot3)\), as according to the Pythagorean theorem). Multiplying those two together gives an answer of \(1.5\sqrt2\).