CPhill

P. S.........there may be some combinatoric that could be found for this, but I couldn't think of any other way to do it!! Anyone else have any ideas??

I think the number of diagonals that can be drawn in any poygon with n vertexes (sides), where n >3, is given by: (n)(n-3)/2.

To see this, note that no diagonal can be drawn from a vertex to itself, nor can any diagonal be drawn to the two neighboring vetexes. So this leaves (n-3) vertexes left. And from each vertex, the same number of diagonals can be drawn. So the total number of diagonals that can be drawn is (n)(n-3). But the diagonal from v_i to v_j is the same diagonal drawn from v_j to v_i, so we must divide this total number by two to prevent "double counting."

So we have...... (n)(n-3) / 2.

Nice pic, Anon. !!

We can write √(x²) as ..."the absolute value of x" =   l x l

And the absolute value of any number is either 0 (in the case of 0, itself), or a positive number.

So let x = a where a > 0    and rearranging your equation, we have

l a l + a = 0

a + a = 0     which is impossible, since adding two postives can never  = 0

So let x = 0

l 0 l + 0 = 0

0 + 0 = 0   ....and x = 0 is OK

Now, let x =a  where a < 0

l a l + a  = 0

-a + a = 0 ....note that, if a < 0, then the absolute value of a is just -a...and this is OK, too. (For example,

l -3 l = 3.........so if a = -3, then -a = 3.

So the answer to your question is that....... x ≤ 0.

The area of the base  = (length of one side of the base)² = 2² = 4 square units

Also, there are 4 triangles, each with a base = 2 and a height = slant height =7

So each has an area = (1/2)(b)(h) = (1/2)(2)(7) = 7

And 4 tmes this = (4)(1/2)(2)(7) = (2)(2)(7)  = 28 sq units

So the "formula" for the surface area = b² + 2(b)(h), where b is one side of the base and h is the slant height of the pyramid

And the surface area of this particular pyramid = (28 + 4) sq units = 32 sq units.

No. In general, unless a prime number in a fraction divides the other number in a fraction, it can't be reduced. Thus, 11 is prime and doesn't divide 100, so that's all we can do.

Simplifying, we have  x^2 +2x - 55 = 0

I can tell this one is going to be "nasty."   Let's use the on-site calculator to solve.

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{55}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{14}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\ {\mathtt{x}} = {\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{14}}}}{\mathtt{\,-\,}}{\mathtt{1}}\\ \end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{8.483\: \!314\: \!773\: \!547\: \!882\: \!8}}\\ {\mathtt{x}} = {\mathtt{6.483\: \!314\: \!773\: \!547\: \!882\: \!8}}\\ \end{array} \right\}$$

Anonymous was not technically incorrect. One answer was, indeed, around 6.5.

Since it's a quadratic, we'll also (usually) have a second different solution, as well.

You are going to flip a coin once and roll a fair, 6-sided die once. What is the sample space for this experiment?

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"Sample Space" refers to the set of all the possible outcomes for some activity.

So we have 

{H1,H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

This makes sense........There are two possibles outcomes for the coin flip and six possible outcomes for the die roll.

So....the number of ways the first thing could happen - 2 - times the number of ways the second thing could happen - 6 -  gives us the total number of outcomes = 12.

what is the answer to 2/3 (x-5)+1/2 x=-3

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Let's get rid of those fractions by multiplying each side through by the common denominator of both 2 and 3, which is 6. So we have

4(x - 5) + 3x = -18   Simplifying, we get

7x -20 = -18    Add 20 to both sides

7x = 2      Divide both sides by 7

x = 2/7

Make sure to check this answer in the original problem!!

how do you do a(2a+3)-a(a+1) I got 3a + a squared + 1 Is that right

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You're close......let's look at it again.

Distributing the "a" across 2a + 3 in the first expression gives us 2a² + 3a

And distributing the "-a" across a + 1 in the second expression gives us -a² - a

So, combining like terms, we have....... a² + 2a

(I think you might have forgotten to distribute the "-a" across the +1 in the second part.)