All Answers 352710 Answers

The above answer is accurate for the LAST 3 digits from the RIGHT. I think the question says the LAST digit from the RIGHT = Sum FIRST TWO digits from the LEFT. If that is accurate, then the number would be:
1000s = 90, 2000s = 80, 3000s = 70, 4000s = 60, 5000s = 50, 6000s = 40, 7000s = 30, 8000s = 20, 9000s =10.
So, the total =90+80+70+60+50+40+30+20+10 = 450 such numbers.

Let\( f(x) = (x+2)^2-5\).
If the domain of \(f\) is all real numbers,
then \(f\) does not have an inverse function,
but if we restrict the domain of \(f\) to an interval \([c,\infty)\),
then \(f\) may have an inverse function.
What is the smallest value of \(c\) we can use here,
so that \(f\) does have an inverse function?

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x+2)^2-5} \\ \\ x &=& \Big(f^{-1}(x)+2\Big)^2-5 \quad | \quad \text{inverse function} \\ x+5 &=& \Big(f^{-1}(x)+2\Big)^2 \\ \Big(f^{-1}(x)+2\Big)^2 &=& x+5 \\ f^{-1}(x)+2 &=& \pm\sqrt{x+5} \\ \mathbf{f^{-1}(x)} &=& \mathbf{-2 \pm\sqrt{x+5}} \\ \hline \\ \sqrt{x+5} &=& 0 \quad | \quad \text{the smallest value of }x \text{ is } c \\ x+5 &=& 0 \\ x &=& -5 \quad | \quad \text{the smallest value is } -5 \\ \\ \hline \mathbf{[c,\infty)} &=& \mathbf{[-5,\infty)} \\ \hline \end{array}\)

another method

\(z = r e^{i\theta}\\ z^8 = r^8 e^{i8\theta}\\ z^8 + \bar{z}^8 = r^8(e^{i8\theta}+e^{-i8\theta}) = 2r^8\cos(8\theta)\\ z = \dfrac{3+i\sqrt{3}}{2}\\ r^2 = \left(\dfrac 3 2\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2 = 3\\ r^8 = \left(r^2\right)^4 = 81\\ \theta = \arctan\left(\dfrac{\frac{\sqrt{3}}{2}}{\frac 3 2}\right) = \\ \arctan\left(\dfrac{\sqrt{3}}{3}\right) = \dfrac \pi 6\\ 8\theta = \dfrac{4\pi}{3}\\ \cos(8\theta) = -\dfrac 1 2\\ 2r^8 \cos(8\theta) = 2 \cdot 81 \cdot -\dfrac 1 2= -81\)

\(6 sed + 8 (120-sed) = 856\\ -2 sed = -104\\ sed = 52\\ suv=120-52=68\)

:) thank you rom!

Simplify
\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8\)( \(i\) is imaginary)

\(\text{Let $z = \dfrac{3 + i \sqrt{3}}{2}$} \\ \text{Let $\overline{z} = \dfrac{3 - i \sqrt{3}}{2}$} \)

\(\begin{array}{|rcll|} \hline z\cdot \overline{z} &=& \left(\dfrac{3}{2}\right)^2 +\left( \dfrac{\sqrt{3}}{2} \right)^2 \\ &=& \dfrac{9}{4} + \dfrac{3}{4} \\ \mathbf{z\cdot \overline{z}} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right) &=& \dfrac{3}{2} + \dfrac{i \sqrt{3}}{2}+ \dfrac{3}{2} - \dfrac{i \sqrt{3}}{2} \\ &=& 2\cdot \dfrac{3}{2} \\ \mathbf{\left(z+\overline{z} \right)} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z+\overline{z} \right)^2 = 3^2 &=& z^2+2\cdot z\cdot \overline{z} + \overline{z}^2 \\ 3^2 &=& z^2 + \overline{z}^2 +2\cdot 3 \\ z^2 + \overline{z}^2 &=& 9-6 \\ \mathbf{z^2 + \overline{z}^2} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z^2+\overline{z}^2 \right)^2 = 3^2 &=& z^4+2\cdot z^2\cdot \overline{z}^2 + \overline{z}^4 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot \left(z\cdot \overline{z}\right)^2 \\ 9 &=& z^4+ \overline{z}^4 + 2\cdot 3^2 \\ z^4 + \overline{z}^4 &=& 9-18 \\ \mathbf{z^4 + \overline{z}^4} &=& \mathbf{-9} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(z^4+\overline{z}^4 \right)^2 = (-9)^2 &=& z^8+2\cdot z^4\cdot \overline{z}^4 + \overline{z}^8 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot \left(z\cdot \overline{z}\right)^4 \\ 81 &=& z^8+ \overline{z}^8 + 2\cdot 3^4 \\ z^8 + \overline{z}^8 &=& 81-2\cdot 81 \\ \mathbf{z^8 + \overline{z}^8} &=& \mathbf{-81} \\ \hline \end{array} \)

\(\left( \dfrac{3 + i \sqrt{3}}{2} \right)^8 + \left( \dfrac{3 - i \sqrt{3}}{2} \right)^8 = \mathbf{z^8 + \overline{z}^8} = \mathbf{-81} \)

You first have to find the equation of the quadratic function, hint use the intercepts.

After you have the function you can take the derivative to get the function f'(x) defining the rate of change of the graph.

You then apply the formula for the average value of a function over an interval which is,

\(f_{avg}(x) = \frac{1}{b-a}\int_{a}^{b} f(x) dx\)

where b=3 a=1 and f(x) = f'(x).

This should then yield the average rate of change for the function over the given interval.

Since 1 - tan² x is the difference of two squares, it factors into (1 - tan x)(1 + tan x).

Since the denominator is tan x + 1 and that is the same as (1 + tan x), those two factors will be "canceled."
The remaining term in the numerator, 1 - tan x, is the simplified form of this fraction.

\(\text{The equation $x^2+ ax = -14$ has only integer solutions for $x$.}\\ \text{ If $a$ is a positive integer, what is the greatest possible value of $a$?}\)

Die Gleichung \( x ^ 2 + ax = -14\)  enthält nur ganzzahlige Lösungen für  x . Wenn  a  eine positive ganze Zahl ist, was ist der größtmögliche Wert von  a ?

\(\color{BrickRed}x^2+ax+14=0\\ x=-\frac{a}{2}\pm\sqrt{(\frac{a}{2})^2-14}\)

\(a\in \mathbb{N^{odd}}\ |\ \{[(\frac{a}{2})^2-14]\} \subset\{squares\}\)

\(a\in \{9;15\}\)

\(x^2+ax+14=0\)

\(a=9;\ x_1=-2;\ x_2=-7\\ a=15;\ x_1=-1;\ x_2=-14 \)

  !

Sent you a friend request, qkddud!! Please accept it!!

Thank you SO MUCH!

\(\text{probably easiest to just use the cheater formula for the vertex and evaluate the expression there}\\ \text{as it's a downward facing parabola that will be the maximum}\\ \text{the cheater formula for the vertex of $ax^2+bx+c$ is $x_v = -\dfrac b {2a}$}\\ r_v = -\dfrac{40}{2\cdot (-5)} = 4\\ f_v = -5(4^2)+40(4) - 12 = -80 +160-12 = 68\)

\(\dfrac 3 x + \dfrac x 3 = b\\ \dfrac{9+x^2}{3x}=b\\ 9+x^2 = 3bx\\ x^2 - 3bx+9 = 0\\ \text{we could use the quadratic equation and do some further solving but we can see that if $b=2$}\\ \text{then we have $x^2-3bx+9 = (x-3)^2$, which provides exactly one solution, i.e. $x=3$}\)

Ok you didn't like the last answer.

(Incidentally why be so rude?  If you feel the answer is incorrect respond back with why.  Don't just silently downvote me)

So let's look at another way to count these.

There are 3 partitions of 6 using 3 numbers 1-4

(4,1,1), (3,2,1), (2,2,2)

These can be assorted among the lanes so that 

(4,1,1) can be done 3 ways

(3,2,1) can be done 6 ways

(2,2,2) can be done 1 way

Now cars must be in order in the lanes.

(4,1,1) will have 6C4 * 2C1 = 30 arrangements.  You pick 4 cars from 6, sort them, put them in their lane.  Then select 1 car from the 2 remaining 

and put it into a lane.

(3,2,1) will have 6C3 * 3C2 = 60 arrangements

(2,2,2) will have 6C2 * 4C2 = 90 arrangements

so there will be 3*30 + 6*60 + 1*90 = 90+360+90 = 540 arrangements

6 different from the last way I solved it.  I'll have to figure out what's the disparity.

Follow the link:

THX YOU SO MUCH!

Express the equation of the line

Beginning with 1000, there are 10+9+8+7+6+5+4+3+2+1 = 55 such numbers.
Each additional 1000 [2000, 3000, 4000.....etc] has exactly the same number. Hence:
55 x 9 = 495 four-digit numbers.

Despite looking like a scientific calculator, it’s not. It’s a computational engine. However, a few years ago it joined the European calculator union, and now it mostly acts like a calculator. With the proper mix of sweet-talk, curses, and button presses, this computational engine will still give high-precision results for some large integer and small decimal equations.

There is a prerequisite for success: you have to be smarter than the computational engine.

This stops most users in their tracks. 

Let \(a_n = \dfrac{10^n-1}{9}\).
Define \(d_n\) to be the greatest common divisor of \(a_n\) and \(a_{n+1}\).
What is the maximum possible value that \(d_n\) can take on?

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{\dfrac{10^n-1}{9}} \\ \hline \mathbf{a_{n+1}} &=& \mathbf{\dfrac{10^{n+1}-1}{9}} \\\\ &=& \dfrac{10^{n}10-1}{9} \\\\ &=& \dfrac{10^{n}10-(10-9)}{9} \\\\ &=& \dfrac{10^{n}10-10+9}{9} \\\\ &=& \dfrac{10\left( 10^{n} -1 \right) +9}{9} \\\\ &=& \dfrac{10\left( 10^{n} -1 \right)}{9} + 1 \\\\ &=& 10\cdot \left(\dfrac{ 10^{n} -1 }{9}\right) + 1 \quad | \quad \dfrac{10^n-1}{9} = a_n \\\\ \mathbf{a_{n+1}} &=& \mathbf{10a_n + 1} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline d_n &=& \gcd( a_n,\ a_{n+1} ) \\ &=& \gcd( a_{n+1},\ a_n ) \quad | \quad a_{n+1} =10a_n + 1 \\ &=& \gcd( 10a_n+1,\ a_n ) \\ && \boxed{\text{Formula: }\gcd(a,b)=\gcd(a-b,b),\ \text{if } a > b } \\ &=& \gcd( 10a_n+1 - 10a_n,\ a_n ) \\ &=& \gcd( 1 ,\ a_n ) \\ && \boxed{\text{Formula: }\gcd(1,a)=1} \\ &=& \mathbf{1} \\ \hline \end{array}\)

The maximum possible value that \(d_n\) can take on is 1

\({If \ n\ is\ a\ constant\ and\ if\ there\ exists \\ a\ unique\ value\ of\ m\ for\ which\ the\ quadratic\ equation\\ x^2 + mx + (m+n) = 0\ has\ one\ real\ solution,\ then\ find\ n. } \)

\( \color{BrickRed}x^2 + mx + (m+n) = 0\\ x=-\frac{m}{2}\pm\sqrt{(\frac{m}{2})^2-m-n}\ |(p-q-formula)|\)

\((\frac{m}{2})^2-m-n\ge 0\\ n \le\frac{m^2}{4}-m\\ \)

\(n\in\mathbb{R}\ |\ n\le\frac{m^2}{4}-m\)

  !

Calculate a normal vector n to the plane through the points
triple A = (1,-2,1), B = (-3,1,4), C = (1,1,2), as in the picture below,
such that if
\(\mathbf{n} = \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}\),
and \(n_1 + n_2 + n_3 = 7\).

\(\text{Let $\vec{v} = \vec{B} - \vec{C}$ } \\ \text{Let $\vec{w} = \vec{A} - \vec{C}$ }\)

\(\begin{array}{|rcll|} \hline \vec{v} &=& \begin{pmatrix}-3 \\ 1 \\ 4 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{v}} &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \\\\ \vec{w} &=& \begin{pmatrix}1 \\ -2 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\1\\ 2\end{pmatrix} \\ \mathbf{\vec{w}} &=& \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\ \hline \end{array}\)

1.

Vector cross product

\(\small{ \begin{array}{|lrcll|} \hline & \vec{n} =\begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix}&=& \vec{v} \times \vec{w} \\\\ & &=& \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} \times \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} \\\\ & &=& \begin{vmatrix}1&1&1 \\ -4&0&2 \\ 0&-3&-1 \end{vmatrix} \\\\ & &=& \begin{pmatrix} 6 \\ -4 \\ 12 \end{pmatrix} \\ \hline n_1+n_2+n_3: & 6-4+12 &=& 14 \quad | \quad \cdot \dfrac{7}{14} \\\\ & 6\cdot\dfrac{7}{14}-4\cdot\dfrac{7}{14}+12\cdot\dfrac{7}{14} &=& 14\cdot\dfrac{7}{14} \\\\ &\mathbf{ 3-2+6 }&=& \mathbf{7} \\\\ & \vec{n} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array} }\)

2.

Vector dot product

\(\begin{array}{|lrcll|} \hline & \vec{n}\cdot \vec{v} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{-4} \\ \mathbf{0} \\ \mathbf{2} \end{pmatrix} &=& 0 \\ (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ \hline & \vec{n}\cdot \vec{w} &=& 0 \\ & \begin{pmatrix}n_1 \\ n_2 \\ n_3 \end{pmatrix} \cdot \begin{pmatrix}\mathbf{0} \\ \mathbf{-3} \\ \mathbf{-1} \end{pmatrix} &=& 0 \\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ \hline (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ \hline \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)& \mathbf{-4n_1+2n_3} &=& \mathbf{0} \\ &2n_3&=&4n_1 \quad | \quad :2 \\ & \mathbf{n_3} &=& \mathbf{2n_1} \\\\ (2) & \mathbf{-3n_2-n_3} &=& \mathbf{0} \\ & \mathbf{n_3} &=& \mathbf{-3n_2} \\\\ & n_3= 2n_1 &=& -3n_2 \\ & 2n_1 &=& -3n_2 \quad | \quad :2 \\ & \mathbf{n_1} &=& \mathbf{-\dfrac{3}{2}n_2} \\\\ (3) & \mathbf{n_1+n_2+n_3} &=& \mathbf{7} \\ & -\dfrac{3}{2}n_2 + n_2 + -3n_2 &=& 7 \\ & -\dfrac{3}{2}n_2 -2n_2 &=& 7 \quad | \quad \cdot (-1 )\\ & \dfrac{3}{2}n_2 +2n_2 &=& -7 \\ & \dfrac{7}{2}n_2 &=& -7 \quad | \quad \cdot \dfrac{2}{7} \\ & n_2 &=& -7\cdot \dfrac{2}{7} \\ & \mathbf{n_2} &=& \mathbf{-2} \\ \hline & n_3 &=& -3n_2 \\ & n_3 &=& -3\cdot (-2) \\ & \mathbf{n_2} &=& \mathbf{6} \\ \hline & n_1 &=& -\dfrac{3}{2}n_2 \\ & n_1 &=& \left(-\dfrac{3}{2}\right)\cdot(-2) \\ & \mathbf{n_2} &=& \mathbf{3} \\\\ & \mathbf{\vec{n}} &=& \begin{pmatrix}\mathbf{3} \\ \mathbf{-2} \\ \mathbf{6} \end{pmatrix} \\ \hline \end{array}\)

I am qkddud

\(If\ n\ is\ a \ constant\ and\ if\ there\ exists\ a\ unique\ value\ of\ m\ for \\ which\ the\ quadratic\ equation\ x^2 + mx + (m+n) = 0\ has\ one \\ real\ solution,\ then\ find\ n. \)

Sorry The not being able to see the whole problem at once bothered me.