All Answers 353061 Answers

If the mean of 6 numbers is 27, their total sum is 6*27=162. 162-23-16-30-33=60 so the sum of the other two numbers is 60. We know the 2 numbers differ by 2 so 2x-2=60 when x is the greater of the 2. That means x is 31 so one number is 31 and the other is 29. Once we remove the number 16, the total sum becomes 146. Therefore the arithmetic mean is 29.2.

:D

Sum of arithmetic sequence

 Sum₃₂ =  n/2 ( 2a₁ + (n-1)d)

             = 32/2 (2a₁  + (32-1)d)

1448    = 16 (2a₁ + 31d)

5th term = 18 = a₁ + 4d    so   a₁ = 18- 4d     use this value of a₁ in the red equation above to calculate 'd'

then  ninth term will be:    5th term + 4d  = 18 + 4(d)

A curve C has equation 𝑦=𝑥^3 −𝑥^2 −𝑥+2

The point P has x-coordinate 2

a) Find dy/dx in terms of x.

b) Find the equation of the tangent to the curve C at the point P.

   The normal to C at P intersects the x-axis at A.

c) Find the coordinates of A.

Hello Guest!

a)

\(𝑦=𝑥^3 −𝑥^2 −𝑥+2\\ \color{blue}\frac{dy}{dx}=3x^2-2x-1\)

b)

\(P(2,y(2))\\ P(2,2^3 −2^2 −2+2)\)

\(P(2,4)\)

\(y'=3x^2-2x-1\\ m=y'(2)\\ m=3\cdot 2^2-2\cdot 2-1\\ \color{blue}m=7\\ E_P(x)=m(x-x_P)+y_P\)

\(E_P(x)=7(x-2)+4\\ =7x-14+4\)

\(E_P(x)=7x-10\)

c)

\(N_{PA}(x)=-\frac{1}{m}(x-x_P)+y_P\\ N_{PA}(x)=-\frac{1}{7}(x-2)+4\\ =-\frac{1}{7}x+\frac{2}{7}+4\)

\(N_{PA}(x)=-\frac{1}{7}x+\frac{30}{7}\)

\(-\frac{1}{7}x+\frac{30}{7}=0\\ -\frac{1}{7}x=-\frac{30}{7} \)

\(x=30\)

\(A(30,0)\)

The coordinates of A are A(30,0)

  !

A.  the first derivative will be   3 x²  - 2x  - 1

B .   find the 'y' coordinate of point 'P' by sub'ing x = 2 into the original equation

          then find the SLOPE at point 'P' by sub'ing in x = 2 into the first derivative equation found in A.

       use this slope to calculate the line equation given point 'P'

         perpindicular to  this slope is   - 1 / (this slope)      

               can you take it from here?   

The maximum value of  - x^2 -8x+12    would occur at x =  -b/2a = -4   

    use this value in the equation to calculate the max value

Similarly , the MINIMUM value of x ^2 -8x+12 would occur at x = -b/2a = 4     calculate the min value by sub'ing in x = 4

5*10^(-4)*3.5*120 = 0.21     via Web2.0 calculator

Thanks Tiggsy.

I have been playing, unsuccessfully with this one. 

Never would have thought of that.   :)

So it is!  I had read it wrong the whole time! 

thanks Alan :)

I do not understand what you assumed but there is often more than one way to interpret these questions.

I think you found a more complicated way to do so.

Here's a more detailed derivation:

I think it is     \((\frac{1}{7})_{base 10}\;=\;(0.\bar{05})_{base6}\)

I used this video to show me how.

Can't say that I fully understand this question, but here's some algebra that might help.

Consider

\(\displaystyle (a-1)^{2}+(b-24)^{2}+(c-38)^{2} ,\)

(geater than or equal to zero),  for all real values of a, b and c.

Expanding each bracket and rearranging,

\(\displaystyle a^{2}+b^{2}+c^{2}-2(a+24b+38c)+1^{2}+24^{2}+38^{2} \geq0, \\ \text{so} \\ a^{2}+b^{2}+c^{2}+2021 \geq2(a+24b+38c), \\ \text{and since } \\ a+24b+38c \geq 2021, \\a^{2}+b^{2}+c^{2} \geq 2(2021)-2021=2021.\)

The other part, a + 16b + 42c, leads to exactly the same result, but how that's to be presented to answer the question, I don't know.

You are right Melody (though your list chooses the higher score for the host rather than the lower score - but the principle is the same!). 

I assumed a large number of potential rounds, with the random allocation being different each time.

You are right.   There is NO maximum.

just replace the x in the g function with   2x^2-3x+4

and then sub 5 for x.

You tell us what you get.   

If \(\omega^3 = 1\) and \(\omega \neq 1\),

then compute \((1 - \omega + \omega^2)(1 + \omega - \omega^2)\).

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{ (1 - \omega + \omega^2)(1 + \omega - \omega^2) } \\ &=& \Big(1 - (\omega - \omega^2)\Big) \Big(1 + (\omega - \omega^2)\Big) \\ &=& 1 - (\omega - \omega^2)^2 \\ &=& 1 - (\omega^2 - 2\omega^3+\omega^4) \quad | \quad \omega^3=1 \\ &=& 1 - (\omega^2 - 2+\omega^4) \quad | \quad \omega^4=\omega^3*\omega=\omega \\ &=& 1 - (\omega^2 - 2+\omega) \\ &=& 3 - (\omega^2 + \omega) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (\omega^2 + \omega)^2 &=& \omega^4 + 2\omega^3 + \omega^2 \quad | \quad \omega^3=1 \\ (\omega^2 + \omega)^2 &=& \omega^4 + 2 + \omega^2 \quad | \quad \omega^4=\omega^3*\omega=\omega \\ (\omega^2 + \omega)^2 &=& \omega + 2 + \omega^2 \\ (\omega^2 + \omega)^2 &=& \omega^2 + \omega + 2 \\ (\omega^2 + \omega)^2- (\omega^2 + \omega) - 2 &=& 0 \quad | \quad \omega^2 + \omega = x \\ \mathbf{x^2-x-2} &=& \mathbf{0} \\\\ x &=& \dfrac{1\pm\sqrt{1-4*(-2)}}{2} \\ x &=& \dfrac{1\pm\sqrt{9}}{2} \\ x &=& \dfrac{1\pm3}{2} \\\\ x_1 &=& \dfrac{1+3}{2} \\ x_1&=& 2 \\ \omega^2 + \omega &=& 2 \qquad \text{no solution!} \\ && \boxed{\omega \ne 1 } \\\\ x_2 &=& \dfrac{1-3}{2} \\ x_2 &=& -1 \\ \mathbf{\omega^2 + \omega }&=& \mathbf{-1} \\ \mathbf{ (1 - \omega + \omega^2)(1 + \omega - \omega^2) } &=& 3 - (\omega^2 + \omega) \\ &=& 3 - (-1) \\ \mathbf{ (1 - \omega + \omega^2)(1 + \omega - \omega^2) }&=& \mathbf{ 4 } \\ \hline \end{array}\)

Let a and b be the solutions to 5x^2 + 11x + 14 = 0. Find 1/a + 1/b

Hello Guest!

\(5x^2 + 11x + 14 = 0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x=\dfrac{-11\pm \sqrt{11^2-4\cdot 5\cdot 14}}{2\cdot 5}\)

\(a= -0.1i(\sqrt{159}+(-11i))\\ b= 0.1i(\sqrt{159}+11i)\)

\(\dfrac{1}{a}+\dfrac{1}{b}=1/(-0.1i(\sqrt{159}+(-11i)))+1/( 0.1i(\sqrt{159}+11i)) \)

\(\dfrac{1}{a}+\dfrac{1}{b}=-0.7\overline{857142}=-\dfrac{11}{14}\) 

  !

https://web2.0calc.com/questions/triangle-problem_14

https://web2.0calc.com/questions/help-is-needed_5

https://web2.0calc.com/questions/pls-help_10037

https://web2.0calc.com/questions/hxlp

the nums are 3,4,5 and i am dissapointed aops user

You might know that 2, 30-60-90 triangles is an equilateral triangle. Now lets name the side of the equilateral triangle $2s$(for simplicity). Now this means the altitude of a single equilateral triangle is $s\sqrt{3}$ that means 2 of these is $2s\sqrt{3}$. This is length AD. Now E is just the midpoint of BD so BE is just $s$. Finally dividing, $\frac{2s\sqrt{3}}{s}=\boxed{2\sqrt{3}}$

What is the maximum value of x^2- 8x + 12?    

That is a parabola that opens upward.  There is no maximum value, it goes on forever.  I guess you could say infinity is the maximum but I'm reluctant to make the mistake of treating infinity as a number. 

You post a lot of 'answers only'  Moo-ster...how about some explanations/derivations once in a while?    You really do not help anyone with only an answer....

Hi Alan,

Could you show me 2 or more possible sets of draws where there is just one unseeded team hosting?

Because I think there is only 1.

Also answered here:

Hi catmg,

Thanks a lot! However the question says that Rob also gives Sydney and Koreb as many glue sticks as each already has, so your answer was incorrect. That's okay though, because your answer helped me understand how to do it, so I think I can solve it by myself now. Thanks! 

92-29=63

92