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The slope of the line segment is (8 - 4)/(-5 - (-3)) = -2, so the slope of the perpendicular bisector is 1/2.  The midpoint is (-4,6), so by point-slope form, the equation fo the line is y - 6 = 1/2(x + 4).  Then y = 1/2*x + 8, so m + b = 1/2 + 8 = 17/2.

Angles OAP and OBP are right angles

∠AOB = 360 - ( 90 + 90 +70 ) = 110º

∠x = ( 180 - 110 ) / 2

Add all the ratios up:

[4 +2 + 3 + 8] =17

The blue marbles =[3 / 17] * 187 = 33 - Blue marbles in the bag.

That is not the answer.

F * R^2=8, 
F * R^5=64, solve for F, R
F = 2 - This is the first term
R = 2 - This the common ratio.
9th term =2 * 2^(9 - 1)
               =2 * 2^8
               = 512 - the 9th term.

\(ab + b^2 - ac -bc\)

\(= b(a+b)b(a+b)\)

\(=\boxed{ (a+b)(b-c)}\)

[- 8 + L]/2 * 20 = 790, solve for L
[-8 + L] * 10 = 790
[-8 + L] = 79
L =79+8 = 87 - The last and 20th term.

Solve (3x - 7)^2 = (x + 4)^2.  I only found x = 11/2.

Hello Guest!

\(\color{BrickRed} (3x - 7)^2 = (x + 4)^2\\ 9x^2-42x+49=x^2+8x+16\\ 8x^2-50x+33=0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {50 \pm \sqrt{2500-4\cdot 8\cdot 33} \over 2\cdot 8}\)

\(x=\frac{50\pm 38}{16}\)

\(x_1=\frac{11}{2}\)

\(x_2=\frac{3}{4}\)

sample

\((3x - 7)^2 = (x + 4)^2\\ (3\cdot \frac{11}{2} - 7)^2 = (\frac{11}{2} + 4)^2\\ 90.25=90.25\)   q.e.d.

\((3x - 7)^2 = (x + 4)^2\\ (3\cdot \frac{3}{4} - 7)^2 = (\frac{3}{4} + 4)^2\\ 22.5625=22.5625 \)     q.e.d.

  !

Pretty good job young person! The only mistake you made was in the letters "ROR". Each permutation begins with one of the 4 letters:E-L-O-R. Beginning with E, "ROR" should appear 3 times with the letters E-L-L.[ELLROR, ELRORL, ERORLL]. Beginning with L, it should appear another 3 times, but because we have 2 L's, it should appear a total of 6 times.  [LELROR,  LLRORE,  LROREL, LERORL, LLRORE, LLEROR].          Beginning with letter "O" is a non-starter, because it sits smack in the middle of "ROR". That leaves the letter R itself. Beginning with R, we should have another 3 combinations with E-L-L. [RORELL, RORLEL, RORLLE].
So, in summary, we should have 12 "ROR" permutations in total.     

Thanks!!!

but that answer is wrong :(

I solved it with Heron's formula! Please ignore this question :)

Hey I'm back yippee

The number of ways to get 3 integers is 6c3=20

A triangle can be constructed so that its 3 sides have the lengths of the 3 selected numbers... but only when the sum of any 2 selected numbers is greater than the third number.

 Cases possible when the longest side of the triangle is 6: 

(6, 5, 4), (6, 5, 3), (6, 5, 2), (6, 4, 3)

Cases possible when the longest side of the triangle is 5:

(5, 4, 3), (5, 4, 2)

Cases possible when the longest side of the triangle is 4: 

(4, 3, 2)

4+2+1=7

So we get 7/20 as our answer.

Yay!

1.)

so there are 46 people voting for 4 people so there are 46*4=184 possible votes 

2.)

it's the same exept we do 46*5=230

These might NOT be the correct answer but they could be the correct answer but they might not be

The coordinates of the intsersection are ((24 + sqrt(455))/12, 31/12) and ((24 - sqrt(455))/12, 31/12).  Then AB = sqrt(455)/6, so the answer is 455/36.

A septagon is a figure with 7 sides. What is the sum of internal angles (in degrees) of a septagon? 

The short answer is 900^o   

The long answer is that the formula for the sum of the interior angles of any polygon is (n – 2) • 180^o 

where n is the number of sides.  So, for a septagon (7 – 2) • 180^o  =  5 • 180^o  =  900^o  

_.

There are four letters in LLRR, so there would be 4! possible different arrangements. However, there are two repeating letters, so we have to divide by 2!2! to account for overcounting: 4!/2!2! = 4*3*2/4 = 6, so there are six possible different arrangements of the word LLRR. 

(If you're asking for a ratio of the permutations of LLRR vs the permutations of ROLLER, then do 6/(6!/6), where 6!/6 accounts for the arrangements of the word ROLLER and 6 accounts for the arrangements of the word LLRR.)

There are three letters in ROR, so there would be 3! possible different arrangements. However, there is one repeating letter, so we have to divide by 2! to account for overcounting: 3!/2! = 6/2 = 3, so there are three possible different arrangements of the word ROR. 

(If you're asking for a ratio of the permutations of ROR vs the permutations of ROLLER, then do 3/(6!/6), where 6!6 accounts for the arrangements of the word ROLLER and 3 accounts for the arrangements of the word ROR.)

I hope I helped!

Well the Big-Test before college

Well High-School is like an important school becfore you go to college I guess? 

The answer is 4.

Since the square and the circle, share teh circle's diameter, the diagonal of the square is also 5. Therefore the length of the side of the square is \(\frac{5}{√2}\). So the area of the square is \((\frac{5}{√2})^2\) which equals \(\frac{25}{2}\).

Next, we figure out the area of the circle. Since the diameter is 5, the radius is \(\frac{5}{2}\). The area is pi multiplied by the radius sqaured.

So it would be 3.14 (In class, my teacher taught us to use 3.14 as pi) times \(\frac{25}{4}\)which equals at the end, 19.625! If it is incorrect, try typing the answer in terms of pi, see if that helps, or I might've done something wrong.