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I is a parabola with (1,-18) as its vertex and opening upwards touches (4,0) which is one of the roots , a straight line y=-x+11 pass through the parabola at P and Q , find the coordinates of P and Q.

$$\boxed{Parabola: y=a(x-x_v)^2+y_v }\qquad x_v=1 \quad y_v=-18\\
a? \quad x = 4 \quad y = 0 \\
0=a(4-1)^2-18\\\\
a=\dfrac{18}{3^2}=\frac{18}{9}=2\\\\
\boxed{Parabola: y=2(x-1)^2-18 \quad line: y=-x+11}$$

The cut:

$$y=-x+11\\
y+x=11 \\
x = 11-y \\
x-1 = 10-y$$

We set x-1 = 10 - y in Parabola:

$$y=2(x-1)^2-18 \\
y=2(10-y)^2-18 \\
y=2(100-20y+y^2)-18\\
y=200-40y+2y^2-18\\
\boxed{2y^2-41y+182=0}\\
y_{1,2}=\dfrac{41 \pm \sqrt{41^2-4*2*182} }{2*2}=\dfrac{41 \pm \sqrt{225} }{4}=\dfrac{41 \pm 15 }{4}\\\\
y_1=\frac{56}{4}=14 \qquad y_2 = \frac{26}{4}=6.5\\
x_1 = 11-y_1 =11-14=-3 \qquad x_2=11-y_2=11-6.5=4.5$$

Point P = (-3, 14) and Point Q = (4.5, 6.5)

6 letters in: 'my hand'

@@ End of Day Wrap   Wed 11 /2/15 Sydney, Australia Time 12:55 am  (Really Thurs morn)  ♪ ♫

Good evening/morning 

Well I have been busy and I know CPhill was equally busy earlier in the day.  Just as well we have so many helpers.  Today these were  Alan, TakahiroMaeda, Heureka, CPhill, Tetration, Wildhorse, JuiceBox, GenericUserNameHere, Saseflower, geno3141, BrittanyJ, Thatguy0123, mathematician, HappyBJ, TBoy1, Sobewo and of course our ever helpful Rosala (who I temporarily forgot, sorry Rosala).  Thanks all, you are all much appreciated. 

Interest posts:

Thurs 12/2/15

Nice, Melody...

I always like these kind of problems...

Thanks, heureka....here's another way to consider this....

If 10 pails fill 5/8 of the tank, than 10 / (5/8)  = 80/5 = 16 pails fill the whole tank.

Then, when 4 more pails are added, 14/16 of the the tank must be filled.

And 5 jugs fill the other 2/16 = 1/8.  So, 5/(1/8)  = 5 * 8  = 40 jugs fill the whole tank....!!!

Thank you Heureka and Alan 

We all make mistakes anon.  Don't sweat it 

I is a parabola with (1,-18) as its vertex and opening upwards touches (4,0) which is one of the roots , a straight line y=-x+11 pass through the parabola at P and Q , find the coordinates of P and Q.

(x-1)^2=4a(y+18)     Passes through  (4,0)

(4-1)^2=4a(0+18)

9=72a

a=9/72=1/8

So the equation of the parabola is

 (x-1)^2=4*(1/8)(y+18)

 (x-1)^2=(1/2)(y+18)

2(x-1)^2=y+18

y=2(x-1)^2-18

y=2(x^2-2x+1)-18

y=2x^2-4x+2-18

y=2x^2-4x-16

Now we need to solve this simultaneously with  y=-x+11

-x+11=2x^2-4x-16

2x^2-3x-27=0

I want two numbers that add to -3 and  mult to 2*-27=2*3*-9=6*-9     The numbers are 6 and -9

replace -3x with 6x-9x

2x^2+6x-9x-27=0

2x(x+3)-9(x+3)=0

(2x-9)(x+3)=0

2x-9=0   or   x+3=0

2x=9      or    x=-3

x=4.5      or    x=-3

Now we need to solve y=2x^2-4x-16 simultaneously with  y=-x+11

When x=4.5   y=-4.5+11=6.5      

check      $${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{4.5}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{4.5}}{\mathtt{\,-\,}}{\mathtt{16}} = {\frac{{\mathtt{13}}}{{\mathtt{2}}}} = {\mathtt{6.5}}$$      well that one works

When x=-3       y=--3+11=3+11=14

check       $${\mathtt{2}}{\mathtt{\,\times\,}}{\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}\left(-{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{16}} = {\mathtt{14}}$$         good that one is correct too

so       P(-3,14)   and    Q( 4.5, 6.5)       [or vise versa]

Beng, Chandra and Danial have some stickers. Chandra and Danial have 7/10 of the stickers. Beng and Danial have 6/7 of the stickers. Beng and Chandra have a total of 620 stickers. How many more stickers did Danial collect than Beng

$$\small{\text{
All Stickers $=x$
}}\\\\
\begin{array}{lcccc}
(1) & b+c+d=x & \quad c+d=x-b &\quad b+d=x-c &\quad b+c=x-d\\\\
\hline \\
(2) & c+d = \frac{7}{10}x &\frac{7}{10}x = x-b \\\\
(3) & b+d = \frac{6}{7}x & &\frac{6}{7}x = x-c \\\\
(4) & b+c = 620 & && 620 = x-d \\\\
\hline \\
& b+c+d=x & b=\frac{3}{10}x & c=\frac{1}{7}x & d=x-620\\\\
\end{array}\\
\begin{array}{rrcl}
& \frac{3}{10}x + \frac{1}{7}x + x-620 &=& x\\ \\
& x &=& 1400
\end{array}$$

$$\\b=\frac{3}{10}*1400 = 420\\\\
c=\frac{1}{7}*1400 = 200\\\\
d=1400-620 = 780\\\\
d-b=780-420 = 360$$

How many more stickers did Danial collect than Beng?  360

i made a mistook.

39/70-3/10=39/70-21/70=18/70

(18/70)*1400=360 ,so Danial collect more 360 than Beng.

i am so sorry guys.

I think we need an adjudicator on this one - I don't have time at present   

Thank you   :))

Beng and Danial have 6/7 of the stickers ,so chandra have 1/7 of the total stickers

so Danial have 7/10-1/7=39/70 of the total stickers , so beng have 6/7-39/70=21/70=3/10

Beng and chandra have 1/7+3/10=31/70,so the Danial ,Beng and chandra have 620/31*70=1400

Danial collect  more 1400*(39/70-3/10)=1400*（9/70）=180 stickers than Beng

Use the calc on the home page  :)

if a and b are the roots of the quadratic equation x^2+4x-k=0 and a>b, find the value of a-b

[I have replace a with $$\alpha$$  and b with   $$\beta$$   because I wanted a and be to stand for the quadratic coefficients]

For the quadratic  

$$\boxed{ax^2+bx+c=0 \;\:If\; the\; roots\; are\; \alpha \;and\; \beta\; \\\;then\;
\alpha+\beta=\frac{-b}{a}\;\;and\;\;\alpha+\beta=\frac{c}{a}}$$

$$\\\alpha+\beta=-4\qquad \alpha\beta=-k\\\\
(\alpha+\beta)^2-4\alpha\beta=\alpha^2+\beta^2+2\alpha\beta-4\alpha\beta=
\alpha^2+\beta^2-2\alpha\beta=(\alpha-\beta)^2\\\\
so\\\\
(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\\\\
(\alpha-\beta)^2=(-4)^2-4*-k\\\\
(\alpha-\beta)^2=16+4k\\\\
\alpha-\beta=\pm\sqrt{16+4k}\\\\
But\; \alpha\; is \;bigger\; than\; \beta\; so\\\\
\alpha-\beta=\sqrt{16+4k}\\\\
\alpha-\beta=2\sqrt{4+k}\\\\$$

f(x)=5x^2-3 when f(t+6)

$$\\f(x)=5x^2-3\\\\
f(t+6)=5(t+6)^2-3\\\\
f(t+6)=5(t^2+12t+36)-3\\\\
f(t+6)=5(t^2+12t+36)-3\\\\
f(t+6)=5t^2+60t+180-3\\\\
f(t+6)=5t^2+60t+177\\\\$$

Using an obvious notation we have:

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This was entertaining.

Factoring quadratics to rap.

10 pails of water can fill 5/8 of a tank. Another 4 pails and 5 jugs of water are needed to fill the tank to its brim. How many jugs of water can the tank hold ?

$$\dfrac{ \frac{5}{8}\ \mathrm{tank} } { 10\ \mathrm{pails}} * 14 \ \mathrm{pails} = \frac{5*14}{8*10} \ \mathrm{tank} = \frac{7}{8}\ \mathrm{tank} \\\\
\frac{7}{8}\ \mathrm{tank} + 5\ \mathrm{jugs} = 1\ \mathrm{tank} \\
5\ \mathrm{jugs} = 1\ \mathrm{tank}- \frac{7}{8}\ \mathrm{tank} =\frac{1}{8}\ \mathrm{tank} \quad | \quad * 8\\
5*8\ \mathrm{jugs} = \frac{8}{8}\ \mathrm{tank} \\
40\ \mathrm{jugs} = 1\ \mathrm{tank} \\$$

Thanks Alan,

I am going to play with this one as well because I  need the practice.

Nikola does   $$\frac{1}{15}$$   of a unit of work in a day.

Nikola and llia together do  $$\frac{1}{10}$$  of a unit of work in a day

so llia must do  $$\frac{1}{10}-\frac{1}{15} = \frac{3}{30}-\frac{2}{30}= \frac{1}{30}$$ 

So by herself llia would take 30 days to complete the task

sin^-1x is the same as arcsinx or asinx.  You can get it (asin) on this calculator by pressing the 2nd button first

.

6 letters 

Thanks sobewo.   

And Welcome to web2.0calc forum 

This can also be written as

2(a-5)      or     2a-10

This clip should help :)

It is written as an imaginary number already.

i is the basis of the whole imaginary number system.     i=sqrt(-1)

I suggest that you study these examples