KnockOut

If A, B, C, D, E all lie on the same circle, and the distances between each consecutive point is the same, you know that these five points constitute a regular pentagon inscribed within a circle. Using this, try to find [ABCDE]

Hint: Try drawing triangles from each side length of the pentagon to the middle of the triangle. You should get five congruent isosceles triangles that you can use to find the area.

Starting with the purple figure,

(4x3) + 2(8x3) + 2(8x4) = 12 + 48 + 64 = 124cm^2

Next the yellow figure,

2(5x10) + 2(5x7) + (10x7) + ((10x7) - (4x3)) = 100 + 70 + 70 + 58 = 298cm^2

Total surface area = 124 + 298 = 422 cm^2

1 + cosx = 1/2,

cosx = -0.5

x = -2/3pi or -4/3pi

B is the answer

First, lets find the coordinates of D and E.

D is the midpoint of A(0,6) and B(0,0)

E is the midpoint of B(0,0) and C(8,0)

The coordinates of D and E are (0,3) and (4,0), respectively.

Since we know the coords of D and E, we can now find the equation for lines AE and CD.

Line AE is a line that passes through the points (0,6) and (4,0).

The slope of AE is (0-6)/(4-0) = -6/4 = -3/2

So AE is a line that has a slope of -3/2 passing through the points (0,6) and (4,0)

y = (-3/2)x + b

6 = 0 + b, 

b = 6

y = (-3/2)x + 6

Do the same process of finding the slope for line CD and plugging in coords to get the equation for line CD:

y = (-3/8)x + 3

Since F is the intersection between the two lines, all we have to do is put the two equations equal to each other and solve.

(-3/2)x + 6 = (-3/8)x + 3

x = 8/3

Plug that x value into any y to get

y = 2

And from there just add 2 and 8/3 to get:

14/3 or 4.66...

A line that is parallel to another line means that the slopes of the two lines are the same.

Since our slope is 1, the equation of our line is in this form: y = x + b

If the line passes through the point (3,2), we can plug the coordinates in to find b.

2 = 3 + b,

b = -1

Therefore, the equation of our line is : y = x - 1

The correct answer is D) The student incorrectly wrote the coefficients of the polynomial.

Even though there is no x^2 term presented, when performing synthetic division you need to write in a 0 for the missing terms.

The division should have looked like:

2  I

__I     1     0     2    -3

                 2     4     12

____________________

          1     2     6     9

for XYZ ~ PQR, the ratio of the sides need to be the same.

XY : PQ = 2 : 3

XZ : PR = 12 : 3(x-2)

12 : 3(x-2) = 2 : 3

6(x-2) = 36,

x-2 = 6

x = 8

The only numbers that can be squared to end in 6 are numbers that end in 4 or 6.

We need to find the possible values of the tens digit, so lets let the tens digit be a variable such as x.

The last two digits of our number will look like this:

10x + 6

or this:

10x + 4

I find it easiest to just plug in values from (1-9) for x and see what you get.

If x is 1, the last two digits will be ...56 or ...96

If x is 2, the last two digits will be ...76 or ...76

If x is 3, the last two digits will be ...96 or ...56

If x is 4, the last two digits will be ...16 or ...36

If x is 5, the last two digits will be ...36 or ...16

.

.

We can keep going all the way to x=9, and you will see that the tens digit will always be an odd number...

Therefore the possible values of the tens digit is (1, 3, 5 , 7, 9)

(4+2i) / (-1+i)

= (4+2i) / (i-1)

(4+2i)(i+1) / (i-1)(i+1)

= 2(i^2)+6i+4 / (i^2)-1

= 6i-2+4 / -1-1

= 6i+2 / -2

= -3i-1

x + y + z = 10,

2x + 2y + 2z = 20

   2x + 2y + 2z = 20

+ -2x - 2y - z = -32

__________________

z = -12

x + y - 12 = 10,

x + y = 22,

3x + 3y = 66

3x - 2y - 24 = -8,

3x - 2y = 16

   3x + 3y = 66

-  3x - 2y = 16

_________________

5y = 50

y = 10

x + 10 - 12 = 10,

x = 12

(x. y, z) = (12, 10, -12)