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Consider:         sqrt(4x + 1) / sqrt(–4x – 1)    

Not Allowed:     denominator equal to zero  

                        negative number under a radical  

Denominator:  

x has to be smaller than –1/4   

that is, larger than |1/4| in the negative direction    

to keep the number under the radical greater than zero      

Numerator:  

However, any x smaller than –1/4 creates a negative number under the radical    

Therefore, there is no solution to this problem;  

that is, there is no value of x that will make the expression well-defined.   

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Let (x,y)  be the center

Since all radiuses are equal  the distance from the center to any of the  points  is equal

So  we can solve this  using (-7,10) and (5,4)

(x + 7)^2  + ( y -10)^2  =  (x - 5)^2  + ( y - 4)^2        simplify

x^2 + 14x + 49 + y^2 - 20y + 100  = x^2 -10x + 25 + y^2 -8y + 16

24x  - 12y  =  -108    ⇒     2x - y  = - 9   ⇒  y= 2x + 9      (1) 

And using  (-7,10) and (15, -10)

(x + 7)^2  + (y-10)^2  = (x -15)^2 + ( y +10)^2

x^2 + 14x + 49 + y^2 - 20y + 100  = x^2 -30x + 225 + y^2 + 20y + 100

44x - 40y  = 176

11x - 10y = 44       (2)

Sub (1)  into (2)

11x - 10 ( 2x + 9) =  44

11x -20x - 90  = 44

-9x = 134

x= -134/9

y = 2(-134/9) + 9  =  -187/9

Center of first circle = (0,0) =  P

Center of second  circle = (1,1)  = Q

Can you take it  from here  ???

Simplify as

x^2 - 14x + y^2 + 20y = -17           complete the square on x, y

x^2 - 14x + 49  + y^2 + 20y + 100  = -17 + 49 + 100

( x - 7)^2  + ( y + 10)^2  = 132

The center is (7, -10)   and the radius =  sqrt (132)  =  = sqrt (4 * 33)  = 2sqrt (33)

Let the height of the upper trapezoid = x    and the height of the lower trapezoid = y

Since the areas are equal

(1/2) (x) ( 1 + EF)  = (1/2)y (2 + EF)

x ( 1 + EF)  = y(2 + EF)

x/y = (2 + EF) / ( 1 + EF)

height = x + y =  (2 + EF) + (1 + EF)  =    3 + 2EF

And twice the area of the top trapezoid =  the area of the whole trapezoid

2 ( 1/2) x (1 + EF)  =  (1/2)(x + y)(1 + 2)

x (1 + EF)  = (3/2)(x + y)

(2 + EF) ( 1 + EF)  =  (3/2)(3 + 2EF) 

2 + 3EF + EF^2  = (3/2)(3 + 2EF)

EF^2 + 3EF + 2 = 3EF + 9/2

EF^2 = 9/2 - 2

EF^2  = 5/2

EF = sqrt (5/2) ≈  1.58

Law of Cosines

6^2  = 7^2  + 5^2  - 2(7)(5)cosACB

[ 6^2 - 7^2  - 5^2 ] / [ -2 (7)(5) ] =  cos ACB

-38 / -70  = cos ACB  =  19/35

sin ACB =  sqrt [ 1 - (19/35)^2 ]  =  sqrt  [ 1 - 361/1225]  = sqrt [ 864]/ 35 = 12sqrt(6) / 35  

1/r =  1/4 + 1/9 + 1/36 + 2sqrt [ 1/36 + 1/144 + 1/324 ]

1/r =  14/36  + 2sqrt [49/1296]

1/r = 14/36 + 2 (7/36)

1/r  = 14/36 + 14/36

1/r = 7/9

r = 9/7

(x-2) (x + 6) ≤ (x-2)(x + 5)

x^2 + 4x -12 ≤ x^2 + 3x -10

x ≤ 2

( -inf, 2 ]

An interior angle of  the n-gon  will measure

360 - 2(60) - 90  = 150°

So

(n-2) / n * 180 =  150

180n - 360 = 150n

30n = 360

n = 360 / 30   =   12

you could say that 1/2*10 is 5 for a, 1/2*100 is 50 for b, and for c, more tests gives a higher accuracy rate, so most likely b - although there is no guarentee, because as you know, low chance doesn't mean impossible!

For convenience, let the volume of M = 20 and N =30    (M : N  = 20 : 30  = 2:3)

If M is is increased by 65%  it now holds   20 (1 + .65)  = 33

So

N must be increased by   ( [ 33 / 30 ]  - 1)  = 1.1 -   1 =  .1  =    10%

AD * AB  =180

The shaded area is a  trapezoid with bases of  AB,  AB/3    and  height AD

Its area =

(1/2)(AD)(AB + AB/3)  =

(1/2)(AD)(4/3)(AB)  =

(2/3) AD * AB  =     (2/3)(180) =  120

Let O be the center of the  circle

Angles  OTX and OUX = 90°

Angle TXU =  angle QXR = 60°

So   TOUX is   a quadrilateral  whose interior angles  sum to  360°

So angle   TOU  =   360 - 90 - 90 -60  = 120°  = minor arc  TU

And angle TAU is an inscribed  angle intercepting the  same  minor arc TU

So TAU measures  (1/2) (120)  = 60°

If all the psychology books are together, then so are all the mathematics textbooks

The books can be  arranged in two ways  (psychology first and  then math, or vice-versa)

And for each of these the math books can be arranged in 4! = 24 ways

And the psychology books can be arranged in 3! = 6 ways

So

6 * 24 * 6  = 864 arrangements

The sophomores can occupy  any one of 5 positions

For each of these they can  be  arranged in 3! = 6 ways

And for each of these the  freshmen can be  arranged in 4! = 24 ways

So

5 * 6 * 24  =   720 arrangements

(23, 29, 31, 37, 41, 43, 47)

He could choose any  pair except the last two

So

C(7,2)  - 1  =   20  different  products

                   (21)(1/7) + (15)(–1/5)   

                           3 – 3  =  0   

Nowhere did it say that the multiple had to be an integer.   

The only problem I see with this is the question, is zero positive?

_.

625x = 9

x = 9/625

So

125 (9 / 625) = (125 * 9) / 625  =    (5 *25 * 9) / (25 * 25 )  =   45 / 25 =  9 / 5

2 / (2 - 3x)

The denominator cannot = 0   so x cannot =  (2/3)

So  domain =  (-inf, 2/3) U (2/3 , inf)

We have a lower power polynomial over a higher power polynomial....in such a case we will have a horizontal  asymptote at y  = 0

So....the range =  (-inf, 0) U (0, inf)

f(x)  = e^x

Proof  

(x, f(x) )  = (0,1)  means that  e⁰   =1

(x, f(x)) = (1,e)  means that e¹ =  e

Let the Blue circle = Blue ribbons

Orange circle = Green ribbons

Purple circle = White ribbons