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i use a website called art of problem solving, in their resource called alcumus they said it was 420... im really confused i dont know 

make a graph of the lines x=3y, x=-3y, y=3x, and y=-3x. you can see that it splits the plane into 8 regions.

HOPE THIS HELPED!

That would be true if E were the center of the circle.....but clearly....that's not the case....angle DEA will subtend an arc > 108°

Let us  let the center of the circle be at F

Drawing radii from F to both D and A  will result in right angles CDF  and  BAF

And ABCDF will form an irregular pentagon.....the sum of the interior angles of this pentagon = 540°

Angles DCB and ABC =  108°

So....we can find angle AFD as

540 - 108 - 108 - 90 - 90  = 144°

But AFD is a central angle in the circle....so....its measure = minor arc AD   =  144°

The ansswer was wrong then.

I answered this one a short time ago.

Lets see.

7*5! = 840 that takes care of rotation

There are 6 axes of symmetry so I think that should be divided by 6

7*5!/6 = 140

420 is definitely NOT correct. 

140 might be correct.

180(n-2), 180*3=540. So, 540/5=108 and the answer is 108.

Nice pic Chris :)

Thank you so much!!!

Thank you, CPhill!

Nice, tertre....!!!

\(\text{The total powers of }3 \text{ in }200! \text{ are}\\ \left\lfloor \dfrac{200}{3}\right\rfloor + \left\lfloor \dfrac{200}{9}\right\rfloor + \left\lfloor \dfrac{200}{27}\right\rfloor + \left\lfloor \dfrac{200}{81}\right\rfloor =97\)

The sum of the roots =   -b / a

So....the sum of the roots = 6

So....we have that

-b/ -2  = 6

-b = -12

b = 12

And the product of these roots  =   c /a

So

(3 + sqrt(5) ) (3 - sqrt(5) )  =   c / -2

9 - 5  =  c / -2

4 = c / -2

-8 = c

So....the function is    y = -2x^2 + 12x - 8

The x coordinate of the vertex is  given by -b / [2a]  =  -12 / [2 * - 2] =  3

So....putting this into the function to find the y coordinate of the vertex we have that

-2(3)^2 + 12(3) -8   =  -18 + 36 - 8 =  18 - 8 = 10

So......the vertex is  (3 ,10)

Hi aNotSoSmartPerson

I just did that.

But you are right I added one that was too big.

There are 11   - just like you said   

HiNTS:

It is increasing when the gradient of the tangent is positive    i.e.       \(\frac{dy}{dx}>0\)

It is decreasing when the gradient of the tangent is negative    i.e.       \(\frac{dy}{dx}<0\)

It is even if it is symmetrical around the y axis, which is the line x=0    SO      \(f(-x)=f(x)\)

It is odd if it is has 180 degree rotational symmetrical around the (0,0), SO      \(f(-x)=-f(x)\)

It is invertible is both the original and the inverse are functions. This happens when:

It is invertible is the gradient is always increaing or always decreasing, 0 is ok too. 

     \(\frac{dy}{dx}\ge0\quad or \quad\frac{dy}{dx}\le0 \qquad \text{ for all real values of x}\)  

Now you can do them yourself :)

236/5 + 236/5^2 + 236/5^3 =57 zeros.

Note: Take the INTEGER part only. Ignore the fractional part.

We know that the area of the shaded region plus the area of circle A equals the Area of Circle B. Since the shaded area: area of the circle, 3:1, we have \(3x+x=64\pi\)\(4x=64\pi, x=16\pi\) . Directly, we \(\)get  \(\pi*r^2=16\pi\), so \(r^2=16, r=4.\) Remember that length can't be negative! Thus, the diameter of circle A is \(4*2=\boxed{8}\) cm.

Thank you!

See the following image :

Let D be the center of the circle

Since BA and BC are tangents meeting radii DX and DY.....then....

Angle BYD = Angle BXD = 90°

And angle YBX = 60°

And XBYD is a quadrilateral.....so the sum of its interior angles = 360

So angle YDX = 360 - 90 - 90 - 60  = 120°

And YDX is a central angle in the circle

And angle YZX is an inscribed angle in the circle intercepting the same arc as angle YDX

So....its measure is (1/2)YDX  = 60°

\(\text{The experimental probability is }\dfrac{1}{25}\\ \text{The theoretical probability is } \dfrac{1}{12}\)

\(\dfrac {1}{12} - \dfrac{1}{25} = \dfrac{25-12}{300}=\dfrac{13}{300}\)

answer (c)

\(\text{The number of flavors is }\\ N=7 + \dbinom{7}{2} = 7+21=28\)

\(\text{The first 7 are the flavors that are two scoops of a given flavor}\\ \text{The 21 are the unordered pairs of two different flavors}\)

Area of a 60 degree sector of a circle with radius 12  =     60/360 pi 12^2     ------------   there are TWO of these

   Minus the area of equalateral triangle                   sqrt3 /4  12^2             -------------- there are TWO of these

60/360 pi 12^2 * 2 -  sqrt3 /4  12^2 *2  =  26.088 sq units

I've tried 3, 6, 9, and 12 and none of them worked xd 

I gave up on the problem and apparently the answer is 11, thanks anyways Melody and Rom.

Solution:

We can construct segments of length 1,sqrt2 ,sqrt5 ,sqrt10 ,2 ,2sqrt2 ,sqrt13 ,3 ,3sqrt2  (these can be obtained systematically by considering all lengths of segments from the top left dot to other dots). The following results can be obtained either by inspection or analysis of slope.

For each of 1, sqrt5, sqrt13, 3, there are zero isosceles triangles having it as its base length.
For sqrt10, there is one.
For each of 2sqrt2, 3sqrt2, there are two.
For each of 2, sqrt2, there are three.
This gives a total of 1 + 2 + 2 +3 + 3 = 11  non-congruent isosceles triangles.

Thanks!

\(\{815,725,635,545,455,365,275,185\}\\ \text{8 possible amounts of change (in pennies)}\)