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@CPhill: You say "let us suppose that  \(a>c>b\)", but what if this isn't true? You show that that if you do suppose it you can achieve the limit, but you haven't proved that you can't do even better by supposing something else. This is the tricky bit!

c= cat weight       then   4c = dog weight

the two weights  total  40     soooo    c + 4c = 40         5c= 40      

         now you can find c= cat weight and calculate 4c=dog weight

Part b:

The answer to both parts really depends on the idea that the circumference of a polygon is at least twice the linear distance between any two vertices. So for the second part you just have to show that

\( |a - b| + |b - c| + |c - d| +\cdots+ |m-n|\ge |a-n|\)

and

\(|n-o| +\cdots+ |x - y| + |y - z| + |z - a| \ge |a-n|.\)

Then

\( |a - b| + |b - c| + |c - d| +\cdots+ |x - y| + |y - z| + |z - a|\ge 2|a-n|.\)

The details are essentially the same as in the previous case.

Vertical angles are equal    soo   7y-92 = y-3       y= 14.833 degrees

All of the angles added should equal 360

7y-92  +  y-3   +  3x-1  + 3x-1  = 360

8y - 95   +  6x -2 = 360

8(14.833) - 95  + 6x  - 2 = 360

x = 56.38 degrees

Here's another way to see this .....

We want to permute  any 2 of the 11 team members....this will give us all the possible pairings of passes between any two team members

So  P (11,2)   =  110 

But since the ball is passed three times between  each of these....then  3 * 110   = 330 possible passes

Short answer: yes. Long answer to come.

(There is no “other team” in this question)      TRUE, GA...I misread the Q !   Thnx

Would part be be 10 as well?

Would part b be 10 as well?

  |a - b| + |b - c| + |c - a| = 20

Let us suppose that  a > c > b      with  a, b and  c  >  0

So...we actually have

( a - b) + (c - b) + (a - c)  = 20    simplify

2a - 2b   = 20

a - b   = 10   and  since  a > b

Then

l a - b  l  = 10

Thanks for the encouragement: I had more or less written this one off as a dud!

The answer looks more complicated than it really is: it's a consequence of the fact that the circumference of a triangle must have at least twice the length of any one side – only in this case the triangle is degenerate (all on one line).

Thanks for answering. I want to look at you answer more thoroughly when I have the time.

It looks interesting :)

Actually (this is the same guy who asked the question) you cant because that if you divide the numbers below zero is actually have higher numbers or answers... But if zero... Example: 8 divided by zero will be infinity (because the lower than one you divide... The higher it will be) but no! Cause of this conclusion: if you divide infinity to zero... You must get one... And is you do (infinity divided by zero) + (infinity divided by zero) is 2... From the easier way... (Infinity + infinity) + (zero + zero) is 1... ITS INSANE... Thats why if you divide ANY NUMBER from zero... You get this message: ERROR

-Guest

Medians AD and BE of triangle ABC are perpendicular. If AD = 15 and BE = 20,

then what is the area of triangle ABC?

not true to scale

\(\begin{array}{|rcll|} \hline A_{\triangle ABC}=A &=& \dfrac{ab\sin{C}}{2} \\ \hline \\ A_{\triangle EDC} &=& \dfrac{\dfrac{ab\sin{C}}{4}}{2} \\\\ &=& \dfrac{\dfrac{ab\sin{C}}{2}}{4} \\\\ &=& \dfrac{A}{4} \\ \hline \\ A_{ABDE} &=& \dfrac{AD\cdot BE}{2} \\\\ &=& \dfrac{15\cdot 20}{2} \\\\ &=& 15\cdot 10 \\\\ &=& 150 \\ \hline \\ A &=& A_{\triangle EDC} + A_{ABDE} \\ A &=& \dfrac{A}{4} + 150 \quad & | \quad \cdot 4 \\ 4A &=& A + 600 \\ 3A &=& 600 \quad & | \quad :3 \\ \mathbf{A} &\mathbf{=}& \mathbf{200} \\ \hline \end{array}\)

The area of triangle ABC is 200

Here's a rough drawing of the triangle:

Since the sum of the angle measures in a triangle is  180°...

m∠BCA  =  180° - 54° - 52°   =   74°

Now look at quadrilateral CDHE.

Since the sum of the angle measures in a quadrilateral is  360°...

m∠DHE  =  360° - 74° - 90° - 90°  =  106°

And since  ∠DHE  and  ∠AHB  are vertical angles, they have the same measure.

m∠AHB  =  m∠DHE  =  106°

Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]
for all $x$ such that $x\neq -1$ and $x\neq 2$.
Give your answer as the ordered pair $(A,B)$.

\(\text{Let $x^2 - x - 2 = (x-2)(x+1)$}\)

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad & | \quad \cdot (x-2)(x+1) \\\\ x + 7 &=& \dfrac{A(x-2)(x+1)}{x - 2} + \dfrac{B(x-2)(x+1)}{x + 1} \quad & | \quad x\ne2,\ x\ne -1 \\\\ x + 7 &=& A (x+1) + B(x-2) \\ \\ \hline x=-1: \quad -1 + 7 &=& A (-1+1) + B(-1-2) \\ \quad 6 &=& 0 -3B \\ \quad 6 &=& -3B \\ \quad -3B &=& 6 \quad | \quad :(-3) \\ \quad B &=& \frac{6}{-3} \\ \quad \mathbf{B} & \mathbf{=} & \mathbf{-2} \\ \hline x=2: \quad 2 + 7 &=& A (2+1) + B(2-2) \\ \quad 9 &=& 3A +0 \\ \quad 9 &=& 3A \\ \quad 3A &=& 9 \quad | \quad :3 \\ \quad A &=& \frac{9}{3} \\ \quad \mathbf{A} & \mathbf{=} & \mathbf{3} \\ \hline \end{array} \)

Answer \((3,-2)\)

^{A passing it to B is different from B passing it to A, right?}

That’s correct. When A passes to B, B is receiving, so that is different from B passing to A.   

oh ok thanks!!!!

\(15c\)

- Daisy

Put the 12 over a line and then put 5t on the bottom.

- Daisy

\(37t - 9.85\)

- Daisy

are you sure? like is that the actual answer if i had to write it on a paper? do i have to use the math symbols? like multiplication or anything?

The quotient of a number \(z\) and \(8 \), minus \(9\).

- Daisy

\(\frac{12}{5t}\)

- Daisy