All Answers 358173 Answers

Find the H.C.F of(x²+y²)(x²+2xy+y²),(x²-y²)³ and (x⁶-y⁶) 

$$\\(x^2+y^2)(x^2+2xy+y^2)\\
=(x^2+y^2)(x+y)^2\qquad (1)\\\\\\
(x^2-y^2)^3\\
=(x-y)^3(x+y)^3\qquad (2)\\\\\\
(x^6-y^6)=(x^3-y^3)(x^3+y^3)\\
=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\qquad(3)\\\\\\
(x^2+y^2)(x+y)^2\qquad (1)\\
(x-y)^3(x+y)^3\qquad (2)\\
(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\qquad(3)\\\\\\
HCF= x+y$$

I think that is correct. 

YES, I want to know where Freddy went.   

$$2^{1/3}=\sqrt[3]{2}\approx 1.2599$$

Well if you google it you will probably find out but with many words with silent letters the letter was not always silent. 

Like knife, the k used to be sounded out.

I'm thinking that at some point in the past the h in honest was most likely not silent.

-----------------------

Mihail, people so sometimes ask things other than maths here.

It is just that we are less likely to be able to help them :)

asin(0.78801)

arc sine is the same as inverse sine

Co=Ca+4      (1)

Co+Ca=28     (2)

Sub 1 into 2

Ca+4+Ca=28

2Ca+4=28

2Ca=24

Ca=12

sub back into (1)

Co=12+4=16

Carla is 12

Connie is 16

I do 9x145=1305. now i want to do 1305/ 9√145

$${\mathtt{9}}{\mathtt{\,\times\,}}{\mathtt{145}} = {\mathtt{1\,305}}$$

1305/9*sqrt(145)

$${\frac{{\mathtt{1\,305}}}{{\mathtt{9}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{145}}}} = {\mathtt{1\,746.031\: \!213\: \!924\: \!882\: \!844\: \!6}}$$

I think that we need the other two heights.  

$${\left(-{\mathtt{3}}\right)}^{{\mathtt{25}}} = -{\mathtt{847\,288\,609\,443}}$$

-1/6t+1=3

$$\\\frac{-1t}{6}+1=3 \\\\
\frac{-1t}{6}=2 \\\\
-1t=12 \\\\
t=-12$$

Well I am going to interpret this as either John or Jack but not both

The number of ways that 4 people can be chosen from 10 is    10C4

Now split hese 10 people into 2 groups. 

Jack and John in the first group and the other 8 people in the 2'nd group.

So we want 1 person from group 1 and 3 from group2

that is    2C1*8C3

therefore P(Jack or john but not both selected)=   

$$\dfrac{^2C_1*^8C_3}{^{10}C_4}=\dfrac{2*56}{210}=\dfrac{56}{105}=\dfrac{8}{15}$$

What is the wavelength in meters of ultraviolet light with ν = 2.50 x 1015 s-1 ?

$$\\\small{\text{Light moves with a speed
$ c \,=\, 299792458\ \frac{\text{m}}{\text{s}} $ }}\\
\small{\text{We denote wavelength by
$\lambda = \mathrm{wavelength}$}}\\
\small{\text{We denote frequency by
$\nu=\mathrm{frequency} $}}\\
\small{\text{Frequency is measured in Hertz
$ = Hz = s^{-1}.$}}\\ \\
\small{\text{
$
\boxed{ \lambda =\dfrac{c}{\nu} }\qquad \nu = 2.50 \cdot 10^{15}\ s^{-1}
$
}}\\\\
\small{\text{
$
\lambda = \dfrac{2.99792458\cdot 10^8\ \dfrac{m}{s} }
{2.50 \cdot 10^{15}\ s^{-1} }
$
}}\\\\
\small{\text{
$
= 1.199169832 \cdot 10^{-7}\ m
$
}}\\
\small{\text{
$
= 119.9169832 \cdot 10^{-9}\ m
$
}}\\
\small{\text{
$
= 119.9\ nm
$
}}$$

Not quite

3.25/100 = 0.0325

The height can be anything. 

Takahiro, if you are travelling in a tardis then you are the time lord. 

And you can travel to whenever and whereever you want. :)

$${\mathtt{1\,999}}{\mathtt{\,-\,}}{\mathtt{127}} = {\mathtt{1\,872}}$$

Almost anon

6(x+5)-(x+5)=6x+30 - x - 5

(3 points for your efforts)

Thanks Takahiro,

I have added this thread to the Sticky Topic Puzzles.   

Hi

A pie is ONLY a pastry

Pi   (without the e) is an irrational number that is approximately equal to 3.14159....

Why are you asking this kind of questins here?

Isn't this website about math only?!

$${\left(\left({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{\mathtt{2.915\: \!5}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\frac{\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{\left({\mathtt{2.097\: \!6}}\right)}^{{\mathtt{2}}}}{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{2}} = {\mathtt{144.402\: \!169\: \!76}}$$

You probably should have brackets somewhere.  

Well 1inch = 2.5400cm

Do $${\mathtt{x}}{\mathtt{\,\times\,}}{\mathtt{2.54}}$$ for inch to cm

You learn it 1st grade of highschool.

It's 0.000000002

A ball rolls on the top of a stairway with a horizontal velocity of 2.0 m/s.

The steps are each 20.0 cm high and 20.0 cm wide.

Which step will the ball hit first ?

$$\\\small{\text{horizontal move:
$\boxed{ x=v_0\cdot t} \; \Rightarrow \; t =\dfrac{x}{v_0} \qquad v_0=2\ \frac{m}{s}$
}}\\\\
\small{\text{vertical move:
$\boxed{ y=\frac{g}{2}\cdot t^2 } \quad t =\dfrac{x}{v_0} \; \Rightarrow \; y(x)=\dfrac{g}{2}\cdot \dfrac{x^2}{v_0^2}
\qquad g=10\ \frac{m}{s^2}$
}}\\\\
\small{\text{$x=y(x) ?:\qquad
x=\dfrac{g}{2}\cdot \dfrac{x^2}{v_0^2} \; \Rightarrow \;
x=\dfrac{2\cdot v_0^2 }{g}
$
}}\\
\small{\text{
$
x=\dfrac{2\cdot (2\cdot \frac{m}{s})^2 }{10\cdot \frac{m}{s^2}} \qquad x=\frac{8}{10}\cdot m = 0.8\ m \qquad 0.8\ m = 4 * 0.2\ m
$
}}$$

The ball will hit first step 4.

 The "orthocenter" of a triangle is defined as the point where the three altitudes intersect.....BTW....in some triangles, this point can actually occur outside the triangle.

This isn't that difficult - just a little lengthy.

We need to find some slopes between these points and then write the equations for the lines representing the altitudes. Where these lines meet is the orthocenter of the triangle.

The slope of the base represented by the side with endpoints (2,3) and (7,3) is easy.... it's 0. So the equation of the altitude from the apex, (5,9), to this base is just x = 5

And the slope of the side with endpoints (5,9) and (2,3)  = 2. And a perpendicular line to this side through point (7,3) will have the equation y - 3 = -(1/2)(x - 7) → y = (-1/2)x + 7/2 + 3 → y = (-1/2)x + 13/2

And when x = 5, y = (-1/2)(5)+ 13/2 = 4   and this is the y coordinate of the orthocenter.

Let's prove that the other altitude also intersects these two lines at y =4

The slope on the side with endpoints (5,9) and (7,3) = -3. And a perpendicular line to this side through (2,3) will have the equation y - 3 = (1/3)(x - 2) → y = (1/3)x - 2/3 + 3 →y= (1/3)x + 7/3

And, when x = 5, we have y = (1/3)(5) + 7/3  = 4....just as we expected......!!!

Then, the y coordinate of the orthocenter is 4 and the orthocenter point is (5,4)

BTW....I have a feeling that at least a couple of people on this site may know a shortcut "formula" for this.....Sorry, I have to do it by "brute force"........