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cos(0)  cancels out  cos(180)

cos(5)  cancels out  cos(175)

...

cos(85)  cancels out cos(5)

This leaves you with cos(90); how much is cos(90)?

cos(355)  cancels out  cos(185)

cos(350)  cancels out  cos(190)

...

cos(275)  cancels out  cos(265)

This leaves you with cos(270); how much is cos(270)?

So, how much is  cos(90) + cos(270)?

good job EP!!!

Thank you! Does anyone know how to do #2 and 3?

Let O be the center of the circle.

Let CD be the chord of length 16

Let AOB be the diameter of the circle that is the perpendicular bisector of the chord; A is on minor arc(CD)

     and D is on major arc(CD).

Let the point of intersection of AOB and CD be point X.

Being a radius, OD = 10.

XD = 8.

Triangle(OXD) is a right triangle.     OX² + XD²  =  OD²  

                                                           OX² + 8²  =  10² 

                                                           OX² + 64  =  100

                                                                   OX²  =  36

                                                                     OX  =  6

OX is the distance from the center to the chord.

[100 / 5]  +  [100 / 25 ] = 24 zeros

1/30   + 1/90   = 1/x    solve for 'x' the amount of time required for both working together

3 / 90  + 1/90  = 1/x

4/90 = 1/x

90/4 = 22.5 minutes = x

c+r =10

d+r=20

d+c=24

d+c+r = ?  

Add the first three equations together to get

2c + 2r + 2d = 54     now divide by 2

c+r+d = 27 kg

log₃(x + 7)  <  log₉(x² + 77)

Using the change-of-base formula:

                 [ log(x + 7) / log(3) ]  <  [ log(x² + 77) / log(9) ]

     [ log(x + 7) / log(3) ] · log(9)  <  log(x² + 77)                [multiply both sides by log(9)]

   [ log(x + 7) / log(3) ] · log(3²)  <  log(x² + 77)                [change 9 into 3²]

[ log(x + 7) / log(3) ] · 2 · log(3)  <  log(x² + 77)               [use property of logs]

                         [ log(x + 7) ] · 2  <  log(x² + 77)               [cancel the log(3) factors]

                             2 · log(x + 7)  <  log(x² + 77)               [rewrite]

                                log(x + 7)²  <  log(x² + 77)                [use property of logs]

                                     (x + 7)²  < x² + 77                        [raisee both sides to power of 10]

                            x² + 14x + 49  <  x² + 77                       [multiply out]

                                   14x + 49  <  77

                                           14x  <  28

                                               x  <  2

Let M and N be the midpoints of AB and CD.

The ratio in which each diagonal is divided is equal to the ratio of the lengths of the parallel sides.

AB/CD = AP/PC = BP/PD = 2/3      The ratio of MP to NP is  2/3 also

If CD = 8  then  AB = 8 * (2/3) = 5.33333333333

NP = 9 / 4 = 2.25 

MP = 2.25 * (2/3) = 1.5

The height of the trapezium is     h = MP + NP = 3.75

Area of the trapezium is            A = h/2 ( AB + CD ) = 25 cm²

The area of the triangle BCP      A_bcp = ( 25 - 13 ) / 2    

AD=4

I know this question has been posted before but none of the given answers are correct.

We use the given points to write an equation in slope intercept form. Using the slope formula, the slope of this line is 2/11. We currently have the equation y=2/11x+b. We plug in one of the points into the equation to find b. Using (8, 7) will be easier.

y=2/11x+b

7=2/11(8)+b

7=16/11+b

b=61/11

So the slope intercept equation we have is y=2/11x+61/11. We then convert it into standard form.

y=2/11x+61/11

-2/11x+y=61/11

2x-11y= -61

A+B+C=2+(-11)+(-61)=2-11-61= -70.

Inversely proportional here means that the value of alpha x beta is -6*12=-72 for any value of alpha and beta. We use this equation and plug in the given value of beta, -2, into the equation.

alpha x beta =-72

alpha x -2 =-72

alpha = 36.

We plug in what is given into the slope formula and solve for y. 

 \(\frac{y-7}{4-(-2)}=-\frac{3}{2}\\ \frac{y-7}{6}=-\frac{3}{2}\\ y-7=-9\\ \boxed{y=-2}.\)

We use the given points to write an equation in slope intercept form. Using the slope formula, the slope is 2. We now have y=2x+b. We can plug in either point into the equation to determine b. We plug in (2, 16).

y=2x+b

16=2(2)+b

16=4+b

b=12

Since b represents the y intercept, then the point intersecting the y axis is (0, 12).

I did the same thing but it said -19 was wrong

Tan 111

Suppose f(x) is a cubic polynomial y = ax^3 + bx^2 + cx + d.  Then

d = 47

a + b + c + d = 32

8a + 4b + 2c + d = -13

27a + 9b + 3c + d = 16

==> f(x) = 46/3*x^3 - 57x^2 + 68/3*x + 47.

The sum of the coefficients is 46/3 - 57 + 68/3 = -19.

First problem: By coordinates 

A = (1,0)

B = (0.924328, 0.471208)

C = (1.222783, -1.147542)

D = (0.924328, -0.471208)

==> angle BCD = 75 degrees

Second problem: 

Triangle(BPS) is similar to Triangle(BAC)     --->     Area(BPS) / Area(BAC)  =  BP2 / BA2  =  22 / 32  =  4/9

     Consequently, Area(APSC)  =  5/9 · Area(ABC)

Similarly, Area(DWT) / Area(DAC)  =  4/9     --->     Area(WACT)  =  5/9 · Area(DAC)

Therefore, Area(APSCTW)  =  5/9 · Area(ABCD)     --->     Area(APSCTW)  =  5/9 · 180  =  20

Cool, didn't expect there'd be any on this site xD

x² + 5 / (x - 3)  =  9 + 5 / (x - 3)

First:  note that x cannot = 3.

Subtract  5 / (x - 3)  from both sides:  x²  =  9 

--->  Either  x = 3  or  x = -3  (but x cannot = 3)

--->   x = -3

an incircle is where the point that the bisectors intersect that forms a circle touching the sides of the triangle 

so we have that OD=4 , BD=6, and DC=8 so BC=14 and we can find the area so they are 14*4/2*2=64 so the area of the pentagon OECBF where O is the incenter and E is where they intersect AC and F is where they intersect AB. We only need to find AEOF which is 2 times AOF so the height of AOF is 4 and the base is x so we get that the area is 2x+64