All Answers 356000 Answers

\(a\mathbf{v} + b\mathbf{w} + c\mathbf{x} = \vec{0}\\ \begin{pmatrix} 1&1&-1\\2&4&6\\1&5&15 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \begin{pmatrix} 1&1&-1\\0&2&8\\0&4&16 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \begin{pmatrix} 1&1&-1\\0&1&4\\0&0&0 \end{pmatrix}\cdot\begin{pmatrix} a\\b\\c \end{pmatrix} = \begin{pmatrix} 0\\0\\0 \end{pmatrix}\\ \text{Let }c = k\\ b +4k = 0\\ \boxed{b = -4k}\\ a - 4k - k = 0\\ \boxed{a = 5k}\\ \dfrac{a - b}c = \dfrac{5k-(-4k)}{k} = 9\)

Can you do it with an ordinary old fashioned calculator?

I can but I am asking guest answerer if he/she can.

4^2012 = 2.2115794807575113474152581518103 x 10^1211

Please answer a question like this with another question to make the aker interact or think a little for themselves.

Maybe the asker might even learn somthing then.

Like you coulod say.

What is    |-4|   ?

and what does this sign     \(\le\)    mean?

What have you done towards answering this question?

Great work CalculatorUser!      

You maths is excellent,

Just for a slightly different presentation I could say:

\(C=2\pi r\\ \frac{C}{2\pi}=r\\ \frac{C+1}{2\pi}=\frac{C}{2\pi}+\frac{1}{2\pi}=r+\frac{1}{2\pi} \)

So if the circumference is increased by 1 unit, the radius is increased by  \(\frac{1}{2\pi}\;units\)

So if the circumference is increased by 1 metre, the radius is increased by \(\frac{1}{2\pi}\;metre\)

And this equals an approximate radius increas of  0.159 metres or approx  16cm

This distance between the 2 concentric circles will always be the same.

 It does not matter how big or small the original circle is.     

A permutation of the numbers (1,2,3,...,n) is a rearrangement of the numbers in which each number appears exactly once. For example, (2,5,1,4,3) is a permutation of (1,2,3,4,5). Let \pi = (x_1,x_2,x_3,---,x_n) be a permutation of the numbers (1,2,3,....,n). A fixed point of \pi is an integer k(1 ≤ k ≤ n) such that x_k=k. For example, 4 is a fixed point of the permutation $(2,5,1,4,3). How many permutations of (1,2,3,4,5,6,7) have at least one even fixed point?

I assume 1824 permutations of (1,2,3,4,5,6,7) have at least one even fixed point.

 1.) 1234567   3 (even fixed points)
 2.) 1234576   2 (even fixed points)
 3.) 1234657   2 (even fixed points)
 4.) 1234675   2 (even fixed points)
 5.) 1234765   3 (even fixed points)
 6.) 1234756   2 (even fixed points)
 7.) 1235467   2 (even fixed points)
 8.) 1235476   1 (even fixed points)
 9.) 1235647   1 (even fixed points)
 10.) 1235674   1 (even fixed points)
 11.) 1235764   2 (even fixed points)
 12.) 1235746   1 (even fixed points)
\(\cdots\)
 998.) 4731265   1 (even fixed points)
 999.) 4735162   1 (even fixed points)
 1000.) 4735261   1 (even fixed points)
 1001.) 4732561   1 (even fixed points)
 1002.) 4732165   1 (even fixed points)
 1003.) 4713562   1 (even fixed points)
 1004.) 4713265   1 (even fixed points)
 1005.) 4715362   1 (even fixed points)
 1006.) 4715263   1 (even fixed points)
\(\cdots\)
 1410.) 6217453   1 (even fixed points)
 1411.) 6274513   2 (even fixed points)
 1412.) 6274531   2 (even fixed points)
 1413.) 6274153   2 (even fixed points)
 1414.) 6274135   2 (even fixed points)
 1415.) 6274315   2 (even fixed points)
 1416.) 6274351   2 (even fixed points)
 1417.) 6275413   1 (even fixed points)
 1418.) 6275431   1 (even fixed points)
 1419.) 6275143   1 (even fixed points)
\(\cdots\)
 1626.) 7261453   1 (even fixed points)
 1627.) 7214563   3 (even fixed points)
 1628.) 7214536   2 (even fixed points)
 1629.) 7214653   2 (even fixed points)
 1630.) 7214635   2 (even fixed points)
 1631.) 7214365   3 (even fixed points)
 1632.) 7214356   2 (even fixed points)
\(\cdots\)
 1812.) 7164325   1 (even fixed points)
 1813.) 7164235   1 (even fixed points)
 1814.) 7164253   1 (even fixed points)
 1815.) 7124563   2 (even fixed points)
 1816.) 7124536   1 (even fixed points)
 1817.) 7124653   1 (even fixed points)
 1818.) 7124635   1 (even fixed points)
 1819.) 7124365   2 (even fixed points)
 1820.) 7124356   1 (even fixed points)
 1821.) 7125463   1 (even fixed points)
 1822.) 7125364   1 (even fixed points)
 1823.) 7123564   1 (even fixed points)
 1824.) 7123465   1 (even fixed points)

Let u, v, and w be vectors satisfying \(\mathbf{u}\bullet \mathbf{v} = 3,\ \mathbf{u} \bullet \mathbf{w} = 4,\ \mathbf{v} \bullet \mathbf{w} = 5 \).
Then what are \((\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w},\ (\mathbf{w} - \mathbf{u})\bullet \mathbf{v},\ (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u}\) equal to?

\(\begin{array}{|rcll|} \hline \mathbf{u}\bullet \mathbf{v} = \mathbf{v}\bullet \mathbf{u} &=& 3 \\ \mathbf{u} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{u} &=& 4 \\ \mathbf{v} \bullet \mathbf{w}=\mathbf{w} \bullet \mathbf{v} &=& 5 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (\mathbf{u} + 2 \mathbf{v})\bullet \mathbf{w} \\ &=& \mathbf{u} \bullet \mathbf{w}+2\mathbf{v} \bullet \mathbf{w} \\ &=& 4+2\cdot 5 \\ &=&\mathbf{ 14 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && (\mathbf{w} - \mathbf{u})\bullet \mathbf{v} \\ &=& \mathbf{w} \bullet \mathbf{v} - \mathbf{u}\bullet \mathbf{v} \\ &=& 5-3 \\ &=&\mathbf{ 2 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && (3\mathbf{v} - 2 \mathbf{w})\bullet \mathbf{u} \\ &=& 3\mathbf{v}\bullet \mathbf{u} - 2\mathbf{w} \bullet \mathbf{u} \\ &=& 3 \cdot 3 -2\cdot 4 \\ &=& 9-8 \\ &=&\mathbf{ 1 } \\ \hline \end{array}\)

If there are vectors \(\mathbf{v} = \begin{pmatrix} 1\\2\\1 \end{pmatrix}\)\(\mathbf{w} = \begin{pmatrix} 1\\4 \\5 \end{pmatrix}\),  and \( \mathbf{x} = \begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix}\).
Find coefficients a, b, and c, not all 0, such that

\(a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\)
and answer with \(\dfrac{a-b}{c}\).

\(\begin{array}{|lrcll|} \hline &\begin{pmatrix} 1&1&-1\\2&4&6\\1&5&15 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (2): &2a +4b+6c &=& 0 \quad | \quad : 2 \\ &a +2b+3c &=& 0 \\\\ &\begin{pmatrix} 1&1&-1\\1&2&3\\1&5&15 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) = (3) - (1) : &\begin{pmatrix} 1&1&-1\\1&2&3\\0&4&16 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) &0\cdot a +4b+16c &=& 0 \quad | \quad : 4 \\ & b+4c &=& 0 \\\\ &\begin{pmatrix} 1&1&-1\\1&2&3\\0&1&4 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (2) = (2) - (1) : &\begin{pmatrix} 1&1&-1\\0&1&4\\0&1&4 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\\\ (3) = (3) - (2) : &\begin{pmatrix} 1&1&-1\\0&1&4\\0&0&0 \end{pmatrix} \cdot \begin{pmatrix} a\\b\\c \end{pmatrix} &=& \begin{pmatrix} 0\\0\\0 \end{pmatrix} \\ \hline \end{array}\)

We set \(c=k\):

\(\begin{array}{|lrcll|} \hline (2): & 0\cdot a+1\cdot b+4 c &=& 0 \quad | \quad c=k \\ & b+4k &=& 0 \\ & \mathbf{ b } &=& \mathbf{ -4k } \\\\ (1): & 1\cdot a+1\cdot b- 1 \cdot c &=& 0 \quad | \quad c=k,\ b=-4k \\ & a-4k-k &=& 0 \\ & a-5k &=& 0 \\ & \mathbf{ a } &=& \mathbf{ 5k } \\ \hline \end{array} \)

proof:

\(\begin{array}{|rcll|} \hline a\begin{pmatrix} 1\\2\\1 \end{pmatrix}+b \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + c\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \quad | \quad a=5k,\ b=-4k,\ c=k \\\\ \hline 5k\begin{pmatrix} 1\\2\\1 \end{pmatrix}-4k \begin{pmatrix} 1\\4 \\5 \end{pmatrix} + k\begin{pmatrix}-1 \\ 6 \\ 15\end{pmatrix} &=& \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \\\\ 5k-4k-k &=& 0 \checkmark \\ 10k-16k+6k &=& 0 \checkmark \\ 5k - 20k+15k &=& 0 \checkmark \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{a+b}{c} } \\\\ &=& \dfrac{5k-(-4k)}{k} \\\\ &=& \dfrac{5k+4k}{k} \\\\ &=& \dfrac{9k}{k} \\\\ &=& \mathbf{9} \\ \hline \end{array} \)

C is the right answer. X can only be between -6 and 2, which none of the other graphs show.

True.

The absolute value of -4 is 4, and 4 is equal to 4

3A  = A ( 1 + .06 / 2)^(N * 2)    where N is in years

3  =  (1.03)^(2N)   take the log of both sides

log 3  = log (1.03)^(2N)

log 3  =  (2N) log (1.03)

N  =  log (3)  [ 2 log (1.03)]   ≈  18.58 years  ≈ 18 years  and 7 months  = 223  months

y = (1.06*0.5)/3

then y roundeddown then multiplied by 6 isyour answer

Correct, CU  !!!

Good job  !!!

\(2{\pi}r=C\)

That is the formula for circumference.

Now to find how far off, we find the difference between the two radii when the concentric circles are formed.

We make a system of equations

\(2{\pi}x=C\)

\(2{\pi}y=(C+1)\)

We have to evaluate \(|y-x|\)

Substituting: \(2{\pi}y-1=2{\pi}x\)

Simplifying: \(1=2{\pi}y-2{\pi}x\)

Solving: \(\frac{1}{2{\pi}}=y-x\)

Solving for \(|y-x|\),

\(\boxed{\frac{1}{2{\pi}}}\)

That was my attempt 

In year 9  the value  is ≈  $1400

In year 14  the value is ≈  $750

The avg rate  of change  =   [  1400 - 750  ]  / [ 9 - 14 ]   =  650 / -5   =  -130 dollars per year

3. What is the maximum value of 4(x+7)(2-x) over all real numbers ?

4  [ 2x + 14 - x^2 - 7x ]  =

4 [ -x^2 - 5x + 14]

-4x^2  - 20x  + 56

The x coordinate of the vertex =   - (-20) / [ 2 * -4]  =  20 / -8   =   -5/2  = -2.5

So.....the max is 

-4 (-5/2)^2 - 20 (-5/2)  + 56   =

-4 ( 25/4)  + 50  +  56  =

-25 + 50  +  56  =

81

Here's the graph : https://www.desmos.com/calculator/pzjfnx8ugd

2.Find the maximum value of  f(x) = -4(x-1)^2+6.

The  vertex  is   (1 , 6)

So....the max value  is 6

You are correct....this graph proves it :  https://www.desmos.com/calculator/ewjtfy8u96

1. The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2+14t-0.4 at time t (in seconds). For how many seconds is the height of the cannonball at least 6 meters?

We want to solve this

-4.9t^2  + 14t - 0.4   = 6        subtract 6 from both sides

-4.9t^2  + 14t  - 6.4  =  0      multiply through by  -10

49t^2 - 140t  + 64   =  0     factor  as

(7 t  - 16)  ( 7t - 4)  =  0

Set each factor to 0  and solve for t  and it will be at 6m   at  4/7 sec  and  16/7 sec

So....it will be a height of 6m  (or above)  for  16/7  - 4/7    =  12/ 7   seconds 

8 * A   =   9 * B  + 1   =   10 * C  +  2

So.....

8A - 1  =   10C + 1

8A - 10C  =  2

4A - 5C  =  1

A       C           B

4       3        Not an integer

9       7        Not an integer

14    11       Not an integer

19    15       Not an integer

24    19       Not an integer

29    23      Not an integer

34    27      Not an integer

39    31      Not an integer

44    35          39

So  he orginally has    8  rows  * 44 per row =  352 trees

When he loses one he  has 9 rows * 39 per row  =  351 trees 

Ad when he loses two he has 10 rows  * 35 per row  =  350 trees    

n mod 8 =0
n mod 9 =1
n mod 10 =2

The LCM of 8/2, 9, 10/2 =180

Using "Chinese Remainder Theorem + Modular Multiplicative Inverse"
180  +  172 = 352 minimum number of trees

What....

And that site unfortunately only gives you the problem, but not a way to solve it

Thanks, guest......that first proof  is  very intuitive  [ with the correct set-up!!!]

Thanks guys! I really want to get that honor roll!