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I think "zeros" are NOT allowed. You have to make a 10 in groups of 3 numbers from 1 to 10.

See the formula used here, under " identical and identical":

https://www.careerbless.com/aptitude/qa/permutations_combinations_imp7.php

Not sure what the answer is....but it has to be more than 8 doesn't it?:

10   0    0

9     1    0

8      2   0

7       3  0

6       4   0

5       5   0

8     1     1

7     2     1

6     2     2

6      3    1

5       4   1

5     3     2

4     3    3

4     4    2

                         fourteen ways?

deleted

This describes an upside down parabola (opens downward...looks like a dome)

so    a^2 = 49   means  a = -7

so you have   -7x^2+bx-6 =0        Using quadratic formula

\(0> {-b \pm \sqrt{b^2-4(7)(-6)} \over -14}\)   yields   b <12.96   and b  > -12.96  

                                                        so b<12.96 is the greatest value

In     Out

5        1

4        2

3        3

2        4

1        5           Five ways  

Hi, \(​​​​​​\lfloor x \rfloor\) that doesn't represent aboslute value, it denotes the greatest integer that is less than or equal to x, otherwise called the "Greatest integer function."

PS, tell me if im wrong, and if it's AoPS, paste answer herre. -LUNAMORT 4 LIFE or Purely Slytherin

HEY CORMIER(123)! i'll be glad to answer. assuming that those weird bracket thingy mean absolulet valuve, you can assume that x has to be positive. then, that means (x)^2=70, which acordding to my caculator, its bout 8.36. HOPE THIS HELPED!

This is a one-step equation. 

\(\hspace{3mm}9h=-36\\ \div 9\hspace{7mm}\div 9\\ \overline{\quad\quad\quad\quad\quad}\\ \hspace{5mm}h=-4\)

Since 9h is equal to -36, any operation performed to both sides of the equation will maintain its equality. In this case, I decided to divide by 9 into both sides to eliminate the coefficient of the lone variable. By doing so, I have effectively isolated and solved for h.

You didn't give the "choices" in your question! However, because it is 45-45-90 right-angle triangle, and the hypotenuse = 24, then the other 2 sides are equal. We can use:

Sin(45) = Opposite side / hypotenuse

Sqrt(2)/2 =opposite side / 24

Opposite side = 24 x sqrt(2)/2

Opposite side =~16.97 and the 3rd side also =~16.97.

So, the 3 sides are =24, 16.97, 16.97.

Your answer is correct because you are given a fraction of a phone in "98.25 phones sold in 10 days".

Now, I would solve it this way:

98.25 phones / 10 days =9.825 phones sold in 1 day.

9.825 phones x 30 days(in April) =294.75, which is what you got. Since you can't have a fraction of a phone, then you simply round it up, in this case, to: 295 phones sold in April.

Total would be   98.25 x 3

Average PER DAY would be   98.25/10

Solve for n:
5 n = 1/2 n (n - 3)
 Write the quadratic polynomial on the right-hand side in standard form.
Expand out terms of the right-hand side:
5 n = n^2/2 - (3 n)/2

Move everything to the left-hand side.
Subtract n^2/2 - (3 n)/2 from both sides:
(13 n)/2 - n^2/2 = 0

Factor the left-hand side.
Factor n and constant terms from the left-hand side:
-1/2 n (n - 13) = 0

Multiply both sides by a constant to simplify the equation.
Multiply both sides by -2:
n (n - 13) = 0

Find the roots of each term in the product separately.
Split into two equations:
n - 13 = 0 or n = 0

Look at the first equation: Solve for n.
Add 13 to both sides:
 n = 13     or      n = 0(Discard) . So this is a 13-gon polygon with 65 diagonals.

Did you see that answer?

I think, when I toss a banana peel it falls toward the Earth, whether I see it or not. I think when Mr. BB makes a post it’s usually bullshit.

(I usually toss... banana peels on a mulch pile, lose screws in a lose screw bin, and bullshit math solutions into the sewer.)

Anyway, the guest did post an answer and then removed it.

Easiest solution, I think, is via co-ordinate geometry.

Align the triangle so that A is at the origin and AC lies along the positive x-axis. (DE will then be perpendicular to the x-axis.)

Let the co-ordinates of B be (4a, 4b) and the co-ordinates of C be (4c, 0).

(The 4's are there so as to avoid fractions later in the calculation.)

Since AB = AC in length, (4a)^2 + (4b)^2 = (4c)^2, so a^2 + b^2 = c^2,

in which case b^2 = -(a^2 - c^2), (needed later).

D is the mid-point of BC so its co-ordinates will be (2a + 2c, 2b), in which case the co-ordinates of E will be (2a + 2c, 0),

and since F is the mid-point of DE, its co-ordinates will be (2a + 2c, b).

Now onto the slopes of the two lines.

The slope of AF will be b/(2a + 2c) and the slope of BE (4b - 0)/(4a - (2a + 2c)) = 4b/(2a - 2c).

The product of these two will be

\(\displaystyle \frac{b}{2(a + c)}.\frac{4b}{2(a-c)}=\frac{b^{2}}{a^{2}-c^{2}}=-1, \)

so the two lines are at right-angles. QED.

The problem can also be solved using vectors and also by a combination of Pythagoras and similar triangles.

Tiggsy.

The guest didn't post his answer, what makes you think that his solution to the question is sloppy?

Yes, Hectictar, great work (as usual). Because of your well-presented solution, and your speed in posting it, you’ve saved the forum from having to bear a yet another pile of chaotic and indeterminate SLOP in the form of Fake Math.

We thank you!

The STARTING midpoint of BG  is at   3, 1   then it is translated 5 to the left

  so the 'x' coordinate becomes   3-5 = -2  

    and translated UP 2 units

       so the 'y' coordinate becomes 1+2 = 3

(-2,3)

Thank you!

Let's write the beginning of the parabola in vertex form

y= a(x-h)^2 +k   (h,k) is the vertex  ...substitute

y= a(x-3)^2 + 2                expand

 y = a (x^2 - 6x +9) + 2         now substitute the point (1,0) in to the equation

0 = a (1^2 -6 +9) + 2

 0 = 4a + 2

so a = -1/2 in THIS equation and it looks like this

y = -1/2 (x^2-6x+9) +2     Simplify

y = -1/2x^2 +3x -4.5 + 2

 y= -1/2 x^2 + 3x - 2.5 

Here is a graph:

1    4    x 3

      2 x 2   x 3

2.

       2 x 13

1      22/55   vs   20/55

2   -18/78     vs   -18/78