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I just spent ages on this and then my page froze and I lost it all.

Here is the short version. I treated it as a co-ordinate geometry question.

Let A(-19,32)  B(-5,12)   C(-22,15)

AB is longer than AC so AB is a diagonal and AC is a side.

The midpoint of AB is ((-12,22)  That is the centre of the square.  I'll call it O

The gradient of AB is  -10/7

so

the gradient of the other diagonal is 7/10 =0.7     (the diagonals are perpendicular)

The distance from the centre O to any vertex is sqrt149

So find the point collinear with C and O that is sqrt149 units from O

( this point is the 4th vertex and it is (-2,29)

So the 4th vertex is    -2+29i

Pictorial representation.

If you have questions then ask away  

Thanks, it does help with b and c but i'm just not quite sure how to tackle a.

(4 hours *43 miles pr. hour + x hours * 59 miles pr. hour) / (4 hours + x hours) = 50

solve for x you get 3.111 ect. = 3.1 hours

The sum converges to 19/36.  Here's one possible derivation of it:

Compute the distance between the parallel lines given by  \(\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)
and \(\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}\).

\(\text{The orthogonal vector of the lines and normalized is }\\ \binom{-3}{4}*\frac{1}{\sqrt{(-3)^2+4^2}}~\text{ or }~\binom{4}{-3}*\frac{1}{\sqrt{3^2+(-4)^2}} \\ \text{because } \binom{4}{3}\binom{-3}{4} = 0 ~\text{ respectively }~\binom{4}{3}\binom{4}{-3} = 0\\ \)

\(\text{Let $ \hat{n} = \frac{1}{5}\binom{-3}{4}$} \)

The distance between the parallel lines is:

\(\begin{array}{|rcll|} \hline d &=&\dfrac{1}{5}\dbinom{-3}{4}\dbinom{-5}{6}-\frac{1}{5}\dbinom{-3}{4}\dbinom{1}{4} \\\\ d &=&\dfrac{1}{5}\Big(15+24-( -3+16 ) \Big) \\\\ d &=&\dfrac{26}{5} \\\\ \mathbf{d} &=& \mathbf{5.2} \\ \hline \end{array}\)

The largest m I could find =5,985

Let
\(\mathbf{v} \text{ and } \mathbf{w} \text{ be vectors such that } \operatorname{proj}_{\mathbf{w}}( \mathbf{v} )= \begin{pmatrix} 4 \\ -7 \end{pmatrix}\).
Find
\(\operatorname{proj}_{-2 {\mathbf{w}}} (3 {\mathbf{v}})\).

\(\begin{array}{|rcll|} \hline \vec{p} &=& \operatorname{proj}_{\mathbf{w}}( \mathbf{v} ) \\\\ &=& \dbinom{4}{-7} \\\\ &=& \dfrac{\vec{w}\vec{v}}{w^2}\vec{w} \\ \hline \vec{P} &=& \operatorname{proj}_{\mathbf{-2w}}( \mathbf{3v} ) \\\\ &=& \dfrac{(-2\vec{w})(3\vec{v})}{|-2w|^2}(-2\vec{w}) \\\\ &=& 6\dfrac{\vec{w}\vec{v}}{2w^2}\vec{w} \\\\ &=& 3\dfrac{\vec{w}\vec{v}}{w^2}\vec{w} \\\\ &=& 3\vec{p} \\\\ &=& 3\dbinom{4}{-7} \\\\ &=& \dbinom{12}{-21} \\ \hline \end{array}\)

\(\operatorname{proj}_{-2 {\mathbf{w}}} (3 {\mathbf{v}}) = \dbinom{12}{-21}.\)

The radius of the sphere is 6*sqrt(5) - 6, so a + c = 5 + 6 = 11.

The formulas for the medians are

AA' = 1/2*sqrt(2*AB^2 + 2*AC^2 - BC^2),

BB' = 1/2*sqrt(2*AB^2 + 2*AB^2 - AC^2),

CC' = 1/2*sqrt(2*AC^2 + 2*BC^2 - AB^2).

Using these formulas, and Pythagoras, we can prove that DE = 8*sqrt(2).

L + T = 34

Five years ago they were five years less old      L- 5     and T- 5

2 ( L - 5)   +  3 (T-5)  = 61              Two equations...two unknowns....solve for L   and   T  

.01 p     +     .05 ( p + 3)   =  .75     solve for p    then n = p+3

15 m/h  * 2 hr   +  x m/h  *  1 hr

________________________                =     14 m/hr

      2  hr   +   1 hr    

(15*2 + x  ) / 3    = 14

30 + x  = 42

x = 12 m/hr

\(0.\overline{1}_b = \\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} \\ \text{let $u = \dfrac 1 b$}\\ \sum \limits_{k=1}^\infty \dfrac{1}{b^k} = \sum \limits_{k=1}^\infty u^k = \\ \dfrac{u}{1-u} = \\ \dfrac{\frac 1 b}{1-\frac 1 b} = \\ \dfrac{1}{b-1} \)

Try choosing the main jurors first.

Projection is linear, so the answer is 2*3*(4,-7) = (24,-42).

1. By stars and bars, the number of ways is C(10,4)*C(9,4) = 26460.

2. Using casework, there are C(7,3) + C(5,3) + C(3,3) = 35 + 10 + 1 = 46 ways to distribute the candies.

40 ft x 70 ft  x 12 ft    = ????   ft³

Stat at  180 degrees      (south)

180   -270 = west       +90

90 -45 =  45 degrees

reverse    45  =180 =225

135 degrees  c c w     225 +135 = 0 degrees   (north)

45 degrees left  (c c w)       45 degrees

spin 180 = 225 degrees

90 degrees left (ccw) 225 +90= 315 degrees   (north east)        

I did not check this answer...you should!

What is the slope-intercept form of the equation of the line that passes through the point (-6, 1) and is perpendicular to the graph of 2x + 3y = -5?

Was ist die Steigungsschnittform der Gleichung der Linie, die durch den Punkt (-6, 1) verläuft und senkrecht zum Graphen von 2x + 3y = -5 ist?

Hello Guest!

\(\color{BrickRed} 2x + 3y = -5\\ y=-\frac{2}{3}x-\frac{5}{3}\\ m_1=-\frac{2}{3}\\ \color{blue}m_2=\frac{3}{2}\)

\(y=m_2(x-x_1)+y_1\\ y=1.5(x+6)+1\)

The slope-intercept form of the equation  that passes through the point (-6, 1) is

\(y=1.5x+10\)

  !

I need help too

2x + 3y = -5    re-arrange

3y = -2x - 5

y  =  -2/3 x  - 5/3       slope = -2/3       perpindicular slope =   -  1  / (-2/3)  = 3/2

y = 3/2 x + b    sub in the given point to find 'b'

1 = 3/2(-6) + b         b = 10

y = 3/2 x  + 10

L = long sleeve        cost of long sleeves =   $ 14  L

T = T shirt = 9- L                 cost of T's     =   $  8 (9-L)         The red equations added together = $ 90

 14L         +       8 (9-L)          =        90            solve for 'L'   the long sleeves, then  T's =   9-L

Why are you so mad, I just simply forgot to state something. I wouldn't have asked this question if I had a calculator.