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thank u 

\(x(x+5) = -n\\ x^2 + 5x +n =0\\ \text{no real roots if discriminant is negative}\\ D = 25-4n\\ D < 0 \Rightarrow n \geq 7\\ P[n \geq 7] = \dfrac{4}{10} = \dfrac{2}{5}\)

This is a perfectly valid question.   

150.7964473723100754   correct to 2 decimal places

First keep only 3 places

=150.796

now if the last digit is 5 or more, which is is because it is 6 then round the second last one up by one

that is the 2nd decimal place goes up from 9 to 10 .... oh dear that is a problem.

So you will have to put thw .78 up to .80

so the answer is 

= 150.80

I've got an account and I do not see a picture either.    

Hi Geno      

Long time, no see.

I hope you've been well.

We have missed you.    And I /we are glad you have put in an appearance.

Why didn't you log in?

\(\hat{u} = B-A\\ d = \dfrac{\|AQ \times \hat{u}\|}{\|\hat{u}\|}\)

\(\hat{u} = (1,2,2)\\ \|\hat{u}\|=3\\ AQ = Q-A=(2,-8,-3)\\ \dfrac{\|AQ \times \hat{u} \|}{\|u\|}= \dfrac{\|(-10, -7, 12)\|}{3} = \dfrac{\sqrt{293}}{3}\\ d=293\)

Only post one question at a time.  It gets confusing for answerers PLUS 

you should be showing evidence of trying to answer these for yourself!

SEE:

https://web2.0calc.com/questions/can-t-figure-this-out_1#r1

First, you graph the triangle. It's going to be a very tall isosceles triangle. Using the fact that the centroid of a triangle splits the median into a 2:1 ratio, and the properties of a 30-60-90 triangle, we can see that the altitude of this triangle is \(3\sqrt 3\). Using the pythagorean theorem, we can see that each of the longer sides is \(\sqrt {82}\). So adding all three sides up, including the one that was already given, we can see that the perimeter of \(2 + 2\sqrt{82}\)

Sorry! I messed up. All the above steps were correct, I just messed up with the Pythagorean theorem. The answer is actually 2 + 4sqrt7

Hope this helps!

Sorry!

A quick lookup finds this:

For a quadratic function in standard form, y=ax2+bx+c , the axis of symmetry is a vertical line x=−b/2a .

Sooo.....

-b/2a = -3

b/a = 6

Of the three dice rolled, one may match the original pair and two are different, and this can occur three ways:  ( 1/6 x 5/6 x 5/6 ) + ( 5/6 x 1/6 x 5/6 ) + ( 5/6 x 5/6 x 1/6 )  =  75/216

[ where 1/6 is the probability of matching the first two and 5/6 is the probability of not matching the first two ]

Or, two may match the original pair and one is different:

                     ( 1/6 x 1/6 x 5/6 ) + ( 1/6 x 5/6 + 1/6 ) + ( 5/6 x 1/6 x 1/6 )  =  15/216

Or, all three may match the original pair:     ( 1/6 x 1/6 x 1/6 )  =  1/216

Or, all three may  be the same, but not match the original pair:

                      ( 5/6 x 1/6 x 1/6 )  =  5/216

[ where 5/6 is the probability of getting a new number and 1/6 is the probability of matching that new number ]

Adding these together:  75/216 + 15/216 + 1/216 + 5/216  =  96/216  =  4/9 

Geno3141

Let $p$ and $q$ be the solutions the the equation $10x^2 - mx + 420 = 0$. We use the fact that the sum and product of the roots of a quadratic equation $ax^2+bx+c = 0$ are given by $-b/a$ and $c/a$, respectively, so $p+q = m/10$ and $pq = 420/10 = 42$. Since $m = 10(p+q)$, we minimize $m$ by minimizing the sum $p+q$. Since $p$ and $q$ are integers and multiply to 42, the possible values of $(p,q)$ are $(1,42),(2,21),(3,14),(6,7),(7,6),(14,3),(21,2),(42,1)$. (Note that if $p$ and $q$ are both negative, then $p+q$ is negative, so $m$ would be negative, which is excluded by the problem.) The sum $p+q$ is minimized when $(p,q) = (6,7)$ or $(7,6)$. In either case, $m = 10(p+q) = 10(6+7) =130

No, it's not. Try again. Look at Melody's solution. 

A number is a perfect square and a perfect cube if and only if it is a perfect sixth power. Note that 10^2=100 and 4^3<100<5^3, while 2^6<100<3^6=9^3. Hence, there are 10 squares and 4 cubes between 1 and 100, inclusive. However, there are also 2 sixth powers, so when we add 10+4  to count the number of squares and cubes, we count these sixth powers twice. However, we don't want to count these sixth powers at all, so we must subtract them twice. This gives us a total of 10+4-2*2=10 different numbers that are perfect squares or perfect cubes, but not both. Thus, our probability is 10/100=1/10. 

HOPE THIS HELPS! 

Thanks LagTho,

Suggesting a graphing calc was a good idea but you have not told guest where to find one or how to use it!

Also a vertex is a co-ordinate, it needs 2 numbers, with brackets, not just one number.

This is the best online graphing calc around 

Guest, you can just put your graph equation straight in on the left hand side.

What is the question?      

You had me confused for a moment,  because I only ever think of them as 1/sqrt2

But guest is correct, they are equivalent.

I never remember the ratios like you obviously do, I just know the triangles that that they come fom.

\(\frac{\pi}{4}\; radians = 45^\circ\\~\\ cos45^\circ=sin45^\circ=\frac{1}{\sqrt2}\)

That makes more sense, its a better explination. But thank you to both of you it really helped me out. 

\((1+x+x^2+......+x^{14})^2\\~\\ =1+2x+3x^2+ .........+15x^{14}+14x^{13}+13x^{12}+............2x^{27}+x^{28}\)

so the product is:

\((1+x+x^2+x^3+.......x^{27})(1+2x+3x^2+ .........+15x^{14}+14x^{13}+13x^{12}+............2x^{27}+x^{28})\\ \)

I am interested in the sum of the  x^28 coefficients.   

This is it

2+3+....+14+15  +14+13+......+2+1

=2(2+3+4+5+6+7+8+9+10+11+12+13+14)+15 +1

= 224

So the coefficient of  x^28=  224

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

Maybe BaldisBasics knows a short cut but he certainly has not explained it to me. 

As an example,

oh right, thanks so much :)

Oh I thought there might have been a simpler way 

Also, to hearts123, this might help answer your question:

\(\frac{\sqrt2}{2}=\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{(\sqrt2)^2}{2\sqrt2}=\frac{2}{2\sqrt2}=\frac{1}{\sqrt2}\)