All Answers 358201 Answers

With one keen observation and a detailed answer for the reals, there is no need for another …unless someone wants to solve for one or more of the complex answers.

Would this post have received this amount of attention without the troll post comments?  Nyet. This post likely would sit here in perpetuity with a wrong answer. 

We know each hour space is 360/12= 30 degrees. When it is 9:20, the hour hand is 1/3 the way between 9 and 10. It is 10 degrees above 9 o clock. Between the hours 4 and 9 is 5 spaces so 5*30=150. The smaller angle in the clock is 150+10=160 degrees. Then the bigger angle is 360-160=200 degrees.

Notice that x-2 and x+1 are factors of x^2-x-2.

\(\frac{x+7}{x^2-x-2}=\frac{a}{x-2}+\frac{b}{x+1}\\ \frac{x+7}{x^2-x-2}=\frac{a(x+1)}{x^2-x-2}+\frac{b(x-2)}{x^2-x-2}\\ x-7=a(x+1)+b(x-2)\\ x-7=ax+a+bx-2b\\\)

ax+bx must add up to the x term on the left side, which is x. a-2b must subtract to the constant on the left side, -7. We then have a system of equations.

a+b=1

a-2b=-7

We find that a= -5/3 and b=8/3.

Here's one way of doing it:

Correct

Even just an approach to my question? Anyone?

x = k/hr

360 / x    -1   = 360 / (x+5)    solve for x

360/x - 360/(x+5)  = 1

360(x+5) - 360x = x (x+5)

1800= x^2+5x

x^2+5x-1800 = 0      Quadratic formula shows x = 40  k/hr   or   -45(throw  out)

so we have (360/x)-(360/(x+5))=60 so we multiply all terms to get 360(x+5)+−360x=1x(x+5) and then −x^2−5x+1800=0 where x=-45 or x=40 but it can't be negative so x=40 and the speed is 40kph

Thanks!!!

x gal = 30 mpg #

50-x  = 15 mpg #

30 x         +      15 (50-x)      = 1050         solve for 'x'  the number of gallons at 30mpg. then the # 15 mpg gallons is 50 -x

100 leaves means 200 pages...what is the next multiple of 7 ?    203

      add 3 more   =   206          pages       is    103 leaves

Looking for the hypotenuse of each right triangle given one leg (53.5 m)  and one angle (64    or   71)

tan 63 = height_b /53.5       solve for height of _building

tan 71 = height_a /53.5       solve for height of _antenna         antenna height is the difference between these two.....

need the diagram  

Well, perhaps not

three squares means the length , l is divided into 3     and each of these segments is the same as the width (because of the square)

perimeter would be   6 of these segements for the sides + 2 more for the ends:

6 (1/3 l) + 2 (1/3 l) = 120

2l + 2/3 l = 120

l = 45   w = 15             three squares would be   15 x 15

pqrs area = 45 x 15  = ......... cm²

Where is the graph?

The tangent of the angle is given by rise/run, so calcuate tan^-1(rise/run)

Since you want the answer in degrees, make sure the calculator you use is in degree mode (or else convert after if it's in radian mode).

How would I calculate a throw with two dice if they both consist out of six sides that are numbered with \(\mathbf{1, 2, 2, 4, 6, 6}\)?
I need to calculate the probability that the sum is greater than 8.
I think the possibilities for the specific situation are (4, 6), (6, 4), (6, 6)
but I'm not sure quite how many possibilities there actually are in total.

\(\mathbf{1, 2, 2, 4, 6, 6}\):

\(\begin{array}{|c|c|c|c|c|c|c|} \hline + &\mathbf{1} & \mathbf{2}& \mathbf{2}& \mathbf{4}& \mathbf{6}& \mathbf{6} \\ \hline \mathbf{1} &2 & 3& 3& 5& 7& 7 \\ \hline \mathbf{2} &3 & 4& 4& 6& 8& 8 \\ \hline \mathbf{2} &3 & 4& 4& 6& 8& 8 \\ \hline \mathbf{4} &5 & 6& 6& 8& \color{red}10& \color{red}10 \\ \hline \mathbf{6} &7 & 8& 8& \color{red}10& \color{red}12& \color{red}12 \\ \hline \mathbf{6} &7 & 8& 8& \color{red}10& \color{red}12& \color{red}12 \\ \hline \end{array} \qquad ~\dfrac{\color{red}8}{36} \)

Normal dice:

\(\begin{array}{|c|c|c|c|c|c|c|} \hline + &\mathbf{1} & \mathbf{2}& \mathbf{3}& \mathbf{4}& \mathbf{5}& \mathbf{6} \\ \hline \mathbf{1} &2 & 3& 4& 5& 6& 7 \\ \hline \mathbf{2} &3 & 4& 5& 6& 7& 8 \\ \hline \mathbf{3} &4 & 5& 6& 7& 8& \color{red}9 \\ \hline \mathbf{4} &5 & 6& 7& 8& \color{red}9& \color{red}10 \\ \hline \mathbf{5} &6 & 7& 8& \color{red}9& \color{red}10& \color{red}11 \\ \hline \mathbf{6} &7 & 8& \color{red}9& \color{red}10& \color{red}11& \color{red}12 \\ \hline \end{array} \qquad ~ \dfrac{\color{red}10}{36} \)

Yes, it will make it much easier for you to visualize it. However, in this attempt you made a small mistake of having a:

7  x  7 table instead of 6  x  6. You added one extra 4 that shouldn't be there. That is why you  have 4 extra 10s.

Thank you so much for the responses! And it's alright, mistakes happen :) So it'd look something like this then? I'll definitely keep it in mind to draw assignments out in the future! 

Thanks answering guest.

Sorry asker guest, I did not read your question properly but you can still use the table only instead of using the headings

1,2,3,4,5,6

jus use

1,2,2,4,6,6

It will work well :)

You can count the number of boxes that make it true and put that over 36 because there are 36 squares (in the grid) in total.

Melody gave you a table for 2 REGULAR 6-sided dice. But yours are NOT normal dice. If you made a table of 36 possible sums of your 2 dice, you would get the following totals:

(2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 10, 10, 10, 10, 12, 12, 12, 12) = 36 sums

Notice the that sums of 10 and 12 are: 4 + 4 = 8. Therefore, the probability of obtaining a sum greater than 8 is:

8 / 36 =2 / 9

Use this table, or one the same to work out all your possibilities.

The top heading row is the number on dice 1

The left heading row it the number on dice 2

You fill out the sums and answer the question/s accordingly.

I do not think your description makes sense.

I take it that X is between A and B

So there is no arc or interval XAB, it would have to be AXB or BXA I suppose.

Can you write your problem so that it makes sense please?

1.

The roots of \(x^2+8x+4\) are the same as the roots of \(Ax^2+Bx+1\). What is \(A+B\)?

The roots:

\(\begin{array}{|rcll|} \hline \mathbf{x^2+8x+4} &=& \mathbf{0} \\\\ x &=& \dfrac{-8 \pm \sqrt{8^2-4*4} }{2} \\ \\ x &=& \dfrac{-8 \pm \sqrt{48} }{2} \\ \\ x &=& \dfrac{-8 \pm \sqrt{3*16} }{2} \\ \\ x &=& \dfrac{-8 \pm 4\sqrt{3} }{2} \\ \\ \mathbf{x_1} &=& \mathbf{-4 + 2\sqrt{3}} \\ \mathbf{x_2} &=& \mathbf{-4 - 2\sqrt{3}} \\\\ \mathbf{x_1x_2} &=& (2\sqrt{3}-4)\left( (-1)(2\sqrt{3}+4) \right) \\ &=& -(2\sqrt{3}-4) (2\sqrt{3}+4) \\ &=& -\left(4*3-16 \right) \\ &=& -\left(-4 \right) \\ \mathbf{x_1x_2} &=& \mathbf{4} \\\\ \mathbf{x_1+x_2} &=& -4 + 2\sqrt{3} -4 - 2\sqrt{3} \\ &=& -4 -4 \\ \mathbf{x_1+x_2} &=& \mathbf{-8} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ Ax^2+Bx+1 } &=& \mathbf{ 0 } \quad | \quad : A \\\\ x^2+ \underbrace{\dfrac{B}{A}}_{=-(x_1+x_2)}x+ \underbrace{\dfrac{1}{A}}_{=x_1x_2} &=& 0 \\\\ \hline \mathbf{ \dfrac{1}{A} } &=& \mathbf{x_1x_2} \quad | \quad \mathbf{x_1x_2=4} \\\\ \dfrac{1}{A} &=& 4 \\\\ \mathbf{ A } &=& \mathbf{\dfrac{1}{4}} \\ \hline \mathbf{\dfrac{B}{A}} &=& -(x_1+x_2) \quad | \quad \mathbf{x_1+x_2=-8} \\\\ \dfrac{B}{A} &=& -(-8) \\\\ B &=& 8A \\ B &=& 8*\dfrac{1}{4} \\ \mathbf{ B} &=& \mathbf{2} \\ \hline A+B &=& \dfrac{1}{4} + 2 \\ \mathbf{A+B} &=& \mathbf{2.25} \\ \hline \end{array}\)