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Holy Guacamole !  I hope you didn't TYPE all of those !      I'd have at LEAST  10 C 3   typos !  

EP is right. There are 10P3 = 720 permutations of the 10 numbers from 0 to 9. Since each permutation begins with one of 10 digits, including zero, then there should be:720 / 10 =72 permutations that will begin with zero. Therefore, the total number of permutations excluding zero at beginning would be: 720 - 72 = 648. Here is a list of ALL numbers that begin with zero. You may count them and should get 72:

{0, 1, 2} | {0, 1, 3} | {0, 1, 4} | {0, 1, 5} | {0, 1, 6} | {0, 1, 7} | {0, 1, 8} | {0, 1, 9} | {0, 2, 1} | {0, 2, 3} | {0, 2, 4} | {0, 2, 5} | {0, 2, 6} | {0, 2, 7} | {0, 2, 8} | {0, 2, 9} | {0, 3, 1} | {0, 3, 2} | {0, 3, 4} | {0, 3, 5} | {0, 3, 6} | {0, 3, 7} | {0, 3, 8} | {0, 3, 9} | {0, 4, 1} | {0, 4, 2} | {0, 4, 3} | {0, 4, 5} | {0, 4, 6} | {0, 4, 7} | {0, 4, 8} | {0, 4, 9} | {0, 5, 1} | {0, 5, 2} | {0, 5, 3} | {0, 5, 4} | {0, 5, 6} | {0, 5, 7} | {0, 5, 8} | {0, 5, 9} | {0, 6, 1} | {0, 6, 2} | {0, 6, 3} | {0, 6, 4} | {0, 6, 5} | {0, 6, 7} | {0, 6, 8} | {0, 6, 9} | {0, 7, 1} | {0, 7, 2} | {0, 7, 3} | {0, 7, 4} | {0, 7, 5} | {0, 7, 6} | {0, 7, 8} | {0, 7, 9} | {0, 8, 1} | {0, 8, 2} | {0, 8, 3} | {0, 8, 4} | {0, 8, 5} | {0, 8, 6} | {0, 8, 7} | {0, 8, 9} | {0, 9, 1} | {0, 9, 2} | {0, 9, 3} | {0, 9, 4} | {0, 9, 5} | {0, 9, 6} | {0, 9, 7} | {0, 9, 8} | {1, 0, 2} | {1, 0, 3} | {1, 0, 4} | {1, 0, 5} | {1, 0, 6} | {1, 0, 7} | {1, 0, 8} | {1, 0, 9} | ... (total: 720)

Thanks so much Omi!

Look at the front face....   it has   2  x  6 squares = 12 squares   .....the BACK face (the one you can't see) has the same

    so far     2  x 12 square

Now look at the side face.  It has   2 x 4 squares = 8 squares    .... the LEFT face (you can't see) does too...

   add         2 x 8 squares

Now look at the TOP face.  It  has  6 x4 squares     so does the BOTTOM face (you can't see)

add      2 *   6 x 4 

Add up the Reds.....what do you get????

Ooops     sorry   misread the question....

My guess

you cannot have   0  then 9 x 8   numbers = 72 numbers

  out of   10 p 3 = 720      720 - 72 = 648 combos  

How many do YOU think it has?

You can just count the lines....or

TWO end triangles have   3 sides (edges) each     2 x 3

then the 'points' of the the two triangles are connected by edges     3

9 edges

You have TWO faces that are   12x 7 cm               2 *  12 x 7

    ONE face that is  4 x 12                                      1 *   4 x 12

and TWO triangle endfaces : area = 1/2 bh           2 *  1/2  4 x 6

Add the reds for your answer...

I'll let you crunch the numbers for an answer

Volume is the number of little cubes (with sides 1*1*1 unit) that fit into the big shape.

The first one has the cubes drawn on it.

Can you see thast there are 6 cubes in the front row ?

And then there are 4 rows going back

So that must be  6+6+6+6 = 24 unit cubes    or   24 cubic units

You could have got this by saying

2*3  * 4 = 24    

See here :)

Cerentie, you have lodge a complaint that people are being rude to you about the number of questions that you have asked.

Perhaps you are unaware of just how many questions you ARE posting.

27 of the last 100 questions are yours!!

How can you possibly be learning from all of these?

We are not here to do your homework for you.

We are not even here to check every one of your answers.

The answerers here want to help people to learn. They are a wonderfully generous and knowledgeable collection of people.

When I see people aske this many question I usually skip over them and help the people who are less demanding.

I know you are not the only offender. There are a couple of others as well, particularly Samjones and Angleray

Actually it is funny because I just realised that the person who you thought was being rude to you was not being rude at all!

They were just amused by your spelling mistake LOL. Their quip was funny :)

The following diagram illustrates the gist of the Fundamental Theorem of Algebra.

 \(P(x)=\underbrace{ax^n+bx^{n-1}+cx^{n-2}+...+yx+z}\\ \hspace{30mm}\text{n complex roots}\)

The degree of the polynomial of \(2x^2+8x+14\) is 2, so the number of complex roots is also 2. 

You can use the quadratic formula to find those roots.

The surface area is the sum of all the areas of the sides.

the origal box has  sides 10cm, 4cm, and 17cm

so the sides are 

10*4=40cm^2                front

10*4=40cm^2                back

10*17=170cm^2             top

10*17=170cm^2             bottom

4*17=68cm^2                side

4*17=68cm^2                side

40+40+170+170+68+68 = 556 cm^2    This is how much wrapping paper that you have,

NOW you have 4 options. Work out the surface area for each one, just like i did for the first one and see which ones you have enough wrapping paper for. 

I just answered this, near enough anyway.....

#1) 

One way to approach this problem is to add up the sides of all the faces. All the faces are rectangular-shaped, so finding their individual area is not too difficult.

We can do the same calculation for the other faces.

Because the above figure is a rectangular prism, the opposite face is equal to one that I already found. 

#2)

A dihedral angle is formed when two planes intersect. \(\angle JAB\) has the same measure because it is included in the dihedral angle. 

Thanks, Melody! 

You got in a mess with this one EP   

But considering how may questions you answer correctly I do not think you need to be upset by a silly error here and there. :)

\(tan \theta = -\frac{3}{2}\\ \)

---

Think about   \(tan \theta = +\frac{3}{2} \)    first

Draw a right angled triangle.

Mark an acute angle as theta.

opp = 3

adj=2

so hypotenuse  =     \(\sqrt{9+4}=\sqrt{13} \)

So for this triangle   \(cos\theta= \frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}\)

----

Now, tan is negative in the 2nd and 4th quadrant

In the 2nd quad Cos is negative and in the 4th quad Cos is positive.

So

\(cos\theta=\pm\frac{2\sqrt{13}}{13}\)

Good job! The answers look good to me!

Yes, it's correct.

AngelRay, my strategy here would be to convert all inequalities to slope-intercept form (\(y=mx+b\)). Converting to slope-intercept form is useful because the form gives one information about how a graph should look. I can, then, compare this information with the four graphs given.

The point-intercept form gives us some information about the corresponding graph of this equation. I will name two of these.

• The slope is 1
• The y-intercept is located at \((1,0)\)

Every graph appears to have the correct slope, but the coordinate of the y-intercept changes from graph to graph. Only the first and last graphs have a y-intercept of \((1,0)\) . Therefore, we have already eliminated two graphs from the list. Notice that the inequality would result in a dashed line. Only the last graph accounts for this technicality, so the bottom-right one is the correct graph.

Yes that is all it means :)

If altitude CD is sqrt3 centimeters, what is the number of square centimeters in the area of triangle ABC?

I prefer to use synthetic division, but I am unsure how to display that in LaTeX. I figured out a way to display the work for long division, though.

\(\hspace{24mm}2x^2+\hspace{2mm}8x+22+\frac{90}{x-4}\\ x-4)\overline{2x^3+0x^2-10x+2\quad\quad\quad\quad}\\ \hspace{4mm}-\left(2x^3-8x^2\right)\\ \hspace{9mm}\overline{\quad\quad\quad\quad\quad}\\ \hspace{23mm}8x^2-10x\\ \hspace{16mm}-\left(8x^2-32x\right)\\ \hspace{20mm}\overline{\quad\quad\quad\quad\quad}\\ \hspace{36mm}22x+2\\ \hspace{29mm}-(22x-88)\\ \hspace{30mm}\overline{\quad\quad\quad\quad\quad\quad}\\ \hspace{48mm}90 \)

I'm a bit confused on how there is a point 3,2. :|