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For a certain value of k, the system has no solutions. What is the value of

\(k x + y + 3z = 10,\\ -4x + 2y + 5z = 7,\\ kx + z = 3 \)

There is no solution, if the determinant \(\begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} = 0\)

\(\begin{array}{|rcll|} \hline \begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} &=& k\begin{vmatrix} 1&3\\2&5\end{vmatrix}+\begin{vmatrix} k&1\\ -4&2\end{vmatrix} \\ &=& k(5-6)+3k+4 \\ &=& -k+2k+4\\ &=& k+4 \quad &| \quad \begin{vmatrix} k&1&3\\-4&2&5\\k&0&1\end{vmatrix} = 0 \\ 0 &=& k+4\\ \mathbf{k} &=& \mathbf{-4} \\ \hline \end{array} \)

Thank you, Melody !

1 No corners cut

2 All corners cut

3 one corner cut

4 two corners cut of narrow edge

5 two corners cut off of long edge

6 two diagonal corners cut

7 three corners cut

\(p=\dfrac 3 8 \cdot \dfrac 5 7 = \dfrac{15}{56}\)

\(\text{Well conceptually it's simple enough}\\ n + d + q = 30\\ 5n+10d+25q=500 \\ n+2d+5q=100\)

\(\text{we can take the difference of these two equations to obtain}\\ d + 4q=70\\ q = \dfrac{70-d}{4},~q \in \mathbb{Z^+}+\{0\}\\ \text{and we can now read off the possible $q$'s}\\ (d,q)=(2,17), (6,16),(10,15),(14,14),(18,13),(22,12), (26,11),(30,10)\\ \text{and of these we select those who's sum is 30 or less}\\ (d,q)=(2,17), (6,16),(10,15),(14,14)\\ \text{and we can then fill in $n$}\\ (n,d,q) = (11,2,17), (8,6,16), (5,10,15), (2,14,14) \)

All sides are equal in a square

x = side length

x * x = area

x*x =64

x = 8 cm       Now use the pythagorean theorem to find the diagonal....

                          a right triangle with two legs , each = 8    and the diagonal is the hypotenuse.    You can do it....

How annoying, now LaTex is not displaying properly ... plus it has  been truncated       

This is an interesting little question.   :)

Zep, the answer is 30 becasue the only numbers that work are 25 and 36.

\((The\;\;number\;\; is \;\;10a+b\\ where\\ 10a+b=\frac{ab}{\frac{a}{b}}\\ 10a+b=b^2\\ 10a=b^2-b\\ 10a=b(b-1)\\ \)

Also

0

The only numbers where b(b-1) is a multiple of 10 is  b=5 and b=6

You can take it from there. 

I've redone some of the latex, I hope it works this time.  I see Heueka has used Latex sucessfully since I last posted.

3x^2 + nx  + 72

The greatest pairwise sum of the factors of 72 is....  72  and 1   = 73....so....we can write

(3x + 1) (x + 72)   =   3x^2 + 1x + 216x + 72  =  3x^2 + 217x + 72

So....the largest value of n  =  217

This is a parabola that turns upward

We can find the x coordinate of the vertex  as  -b/2

So....we can solve this for "b"

(-b/2)^2  + b(-b/2) + 15  < 0

b^2/4  - b^2/2  + 15 < 0     multiply through by 4

b^2 - 2b^2 + 60 < 0

-b^2 + 60< 0   multiply through by - 1  and reverse the inequality sign

b^2 -60 > 0

(b + √60) (b - √60) > 0

We have three possible solution intervals

(-infinity, -√60) or ( -√60, √60)  or ( √60, infinity)

Testing z = 0 in the middle interval in the original inequality makes it untrue

So.....the number of integers between -√60  and √60  =  -7  and 7 inclusive  =  15 integers 

So....15 integer values for b produce no real solutions for z

SERIOUSLY !   TWELVE  THIRTEEN posts so far?   Are you just trying to get your homework done for you?   Have you TRIED to answer ANY of these questions yourself before dumping them on the forum?

x ( x + 7.5) > 38.5

x^2 + 7.5x > 38.5

x^2 + 7.5x - 38.5 > 0    multiply through by 2

2x^2 + 15x - 77 > 0     factor

(2x  - 7) ( x + 11) > 0    set  each factor to 0

2x - 7  = 0             x + 11 = 0

2x  = 7                   x  = - 11

x  = 7/2

We have three possible intervals that make the original inequality true

(-infinity, -11)  or  ( -11, 7/2)  or  (7/2, infinity)

Testing 0  in the middle interval makes the inequality false

So....the two intervals that make the original inequality true are

(-infinity, -11) U ( 7/2, infinity )

the answer is 11 < x < 3.5

[ ( 3x + 2y)^2 ( 3x - 2y)^2 ]^3  =

 ( [ (3x + 2y) (3x - 2y) ]^2 ) ^3  =

[ (3x + 2y) (3x - 2y) ]^6 =

[ 9x^2 - 4y^2 ]^6 

Will have 7 terms

not what i meant!

i  meant that the latex is held with the

\(

plus the -4

then the \)

to look like this:

For what real values of \(x\) is \(-4\) satisfied? Express your answer in interval notation.

The pathway will consist of 4 squares each with a side of x, two rectangles with dimensions of 8 *x   and two rectangles with dimensions of 11*x ......  where x is the width of the path

4x^2  + 2(8*x)  + 2(11*x)  = 120

4x^2 + 16x + 22x  = 120

4x^2 + 38x  = 120

4x^2 + 38x - 120  = 0   factor

(x + 12)(4x - 10)= 0      set each factor to 0 ans solve for x

x + 12  = 0                 4x  - 10  = 0

x = -12  (reject)          4x  = 10

                                    x  = 10/4  = 5/2  =  2.5 ft

So.....the path width is 2.5 ft

Let  x  be the width of the border......then one dimension of the picture and frame  =  ( 10 + 2x)

And the other dimension of the picture and frame  = ( 14 + 2x)

So....the area  =

(10 + 2x) ( 14 + 2x)  = 221      simplify

140 + 28x + 20x + 4x^2  = 221

4x^2 + 48x + 140  = 221       subtract 221 from both sides

4x^2 + 48x  - 81  = 0       factor

(2x - 3) (2x+ 27)  = 0        set each factor to 0   and solve for x

(2x - 3)  = 0                         2x  + 27  = 0

2x  = 3                                  2x  = -27

x  = 3/2  = 1.5 inches            x  = -27/2  = - 13.5  reject

So.....the width is 1.5 inches

x = width of frame

original height 10    new height   10 + 2x

original width   14    new width    14 + 2x

and

(10+ 2x)(14+2x) = 221

4x^2+48x -81 = 0       Use quadratic Formula to find  x = 1.5 inch width of frame

If +- 4 is what that is then

     -4< x < 4

he was trying to do a latex short cut witch is 

\(

and 

\)

which would be

\(

+   -4

+   \)

Sorry...I am unclear on the notation....    \(          What does that mean?