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The radius of the circle  =   5

The equation of the  circle is   x^2 + y^2   = 25    (1)

Re-write  the equation of the line  as    y =(-3/4)x + 7/4    (2)

Sub (2) into (1)

x^2  +  ( 7/4  -(3/4)x)^2    =   25

x^2   + (1/16)(3x - 7)^2 =   25

16x^2  +  9x^2 - 42x + 49   =  400

25x^2  - 42x  -351   = 0

Solving this for  x  gives   x  = -3    and x  = 117/25

And we  only needt to find one associated value  for  y  because AP  = BP

So  A   = (-3  , (-3/4)(-3) + 7/4)   =  (-3, 4)

And AP  =   sqrt  [ ( -3 - 3)^2  +  (4 - 4)^2  ]  =    sqrt [36]   =  6

And OP   = 5   and OA  = 5

So semi-perimeter of  OAP  =  [ 5 + 5 + 6] / 2    = 8

So....using symmetry  [AOBP ]  =

2sqrt   [ 8 * ( 8 - 5)^2  *  (8 - 6)  ]   =

2 sqrt  [ 8 * 9 * 2 ]   = 

2sqrt [ 144 ] = 

2 * 12   =

24 

Lol, this was back then when you rearranged the lines, I forgot I think you

|+|= you put the equal sign as the top and bottom so it makes this

 _

|±|

Angle PBC = 63 degrees and angle BPC = 42 degrees.

I'm sorry but that answer is incorrect

A circle is centered at O  and has an area of 48pi. Let  Q and R  be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO  and triangle PQR  is equilateral, then find the area of triangle PQR.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

OQ = OR = r = 4√3

∠QOR = 30º         ∠QOP = 15º           ∠QPR = 60º

QR = 2 (OQ * sin∠QOP) ==>         QR = 6*sqrt(3)

Height of ΔPQR      h = sqrt[PQ2 - (QR/2)2] = 12*sqrt(3)

[PQR] = h * QR / 2 = 8*sqrt(3)

Let   FI  =  3    IH  = 4    FH   = 5     EG  = height of rectangle  = AB

sin IFH   = 4/5   =sin EFG

sin EFG / EG  =  sin G  / 4

(4/5) / EG  =sin 90 / 4

(4/5)/ EG  = 1/ 4

EG  = (4/5)  /(1/4)  =  16/5   =  AB 

AB = 4 * cos^-1(3/4) = 3.2

We have  to  "cancel"  x's in the expansion

This will  happen  when we  have  the term

C (8,4)  (2x)^4  * ( k/x)^4    = 700000    simplify

70 * 2^4 * k^4   =  700000   divide out 70

70 * 16 * k^4  =  10000

16 * k^4 = 10000 

k^4  =  10000/16    take the positive 4th root

10/2  = k   =  5

This can't happen.....

If one CD is in the wrong case, then there must be (at least) two CDs in the the  wrong case

For example.....if the  correct CDs  are in cases "3" and  "4",  but CD "1" is in case "2"....then CD "2"  must be in case "1"...so....two CDs are in the wrong cases

Couldn't it also be sqrt(2i)?

Edit: Oh wait, never mind, because sqrt(2i) = sqrt((1+i)^2)

Hi guest, sorry for answering so late:

By the angle bisector theorem, QX / QP = RX / RP, and as we know RP and PQ, we can write QX / 9 = RX / 10. However, we also have that QX + RX = 17, so QX = 17 - RX. Thus, (17-RX)/9 = (RX)/10, and 170-10RX = 9RX, thus yielding RX = 170/19 and QX = 153 / 19.

Now, as PRX and QPX share an altitude from the same vertex, the ratios of their areas are the ratios of their bases. Thus, we can compute the area of PQR by heron's formula, which is 36. Thus, [PRX] / [PQX] = RX / QX --> 170/19 * 19/153 = 170/153 = 10/9. We also have that [PRX] + [PQX] = 36, so we can say that [PQX] = 36 - [PRX], so [PRX] / (36 - [PRX]) = 10/9, so 9[PRX] = 360-10[PRX], and [PRX] = 360/19.

Alternatively, we could have dropped an altitude and wrote equations to figure out the area of RPX.

We did all this to find XY. As XY is an altitude of PRX, we have (XY)(PR)/2 = 360/19, and as PR = 10, XY(5) = 360/19, so XY = 72/19

There are several different approaches to this problem, it's up to you to find which one works best!

Area of a circle = pi r^2    

Area of larger circle   minus   area of smaller circle = ring area

pi (10^2) - pi(6^2) = ring area

64pi = ring area 

201.06 units²

The common ratio between 2 terms in this geometric series is 1/18, and the first term of this geometric series is 2.

The sum of an infinite geometric series, if the common ratio is strictly greater than -1 and less than 1 and not equal to zero, is equal to \(a\over(1-r)\), where "a" is the first term and "r" is the common ratio.

Therefore, the answer to this is:

\(\frac{2}{1-\frac{1}{18}} = \frac{2}{\frac{17}{18}} = \frac{36}{17}\)

Common ratio    2 * r = 1/9         r = 1/18

sum = a1 (1-r^n)/(1-r)

       = a1(1/(1-r) ) =   2 (1/(1-1/18)) = 2.12 

.5 in/hr   x 3 hr = 1.5 in added to  11 inches of existing base = 12.5 inches of snow

Times when busses are back at the depot

20     40     60       80     100          120       140 

   35                70                   105               140

They will both be back at the depot at the same time at 140 min ...2 hrs 20 min past their start....  8:20 AM

The question gets cut off at "bus B takes 35 minutes to complete its ro"

72_8  +  77_8  = 162_8

7_8    -   2_8   = 5_8

Note: The first base I tried was base 8. I converted 162_8 to base 10 = 114. Then I tried 2 numbers that add up to 93. I tried 45 + 48 = 93. Then I converted 45 to base 7 =63. And similarly, converted 48 to base 7 = 66. Then added 72 + 77 (base 8) =162_8. That is how I arrived at the answer above.

Hi, 

there are two ways to solve this question

Strategy 1): Divide both sides of the equation by 23

{(128–y) x 23}/23 = 2346/23

Simplify: 128–y = 102

Therefore, y=26

Strategy 2): Expand the left side of the equation

(128–y) x 23 = 2944–23y = 2346

Simplify: –23y=2346–2944

Simplify: –23y=–598

Therefore, when both sides are divided by –23, 

y=26

I hope this helps you :)

thank you!

thank you!

I'd do it like this

\(4x^2 + 14x + c\\ =4(x^2+\frac{14}{4}x+\frac{c}{4})\\ =4(x^2+\frac{7}{2}x+\frac{c}{4})\\\)

now if this is a perfect square then:

\(\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\ \frac{c}{4}=[\frac{7}{4}]\color{red}^2\\ \frac{c}{4}=\frac{49}{16}\\ c=\frac{49}{4}=12\frac{1}{4}\)

LaTex:

\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\
\frac{c}{4}=[\frac{7}{4}]\color{red}^2\\
\frac{c}{4}=\frac{49}{16}\\
c=\frac{49}{4}=12\frac{1}{4}

I split it into 2 possibilities and added the probs at the end.

4hearts,                heart,    a two

\(\frac{1}{52}*\frac{12}{51}*\frac{4}{50}\)

or

4 not of hearts,    heart,     a two.

You can do this one yourself.     

Find all complex numbers z such that z^4 = 4.

Since it is a power of 4, there will be 4 solutions.

one of those solutions will be  \(\pm \sqrt[4]{ 4}=\pm \sqrt{2}\)

the other 2 will be   \(\pm \sqrt[4]{ 4}i=\pm \sqrt{2}\;i\)

"A second equilateral triangle is formed having sides that are of the length of the sides of the first triangle."

I understand that this way: 

1st triangle has a side of 2 units

2nd   .................... side of 6 units

3rd    .................... side of 18 units

and so on........

If it's so, then the percentage increase will be 1,968,300 %