CPhill

5 / [x^2 + 6x + 8]   =  A / [x + 2 ]  + B / [ x + 4]

On the right, we can get a common denominator   and write

[ A ( x + 2)] + [ B(x + 4)]    =  5

(A + B)x + 2A + 4B  = 5

Equating co-efficients, we have

A + B  = 0     →   B = -A   (1)

2A + 4B  = 5     (2)

And subbing (1)  into (2), we have

2A + 4(-A)  = 5

-2A  = 5       →  A  = -5/2  = -2.5

And, using (1), B = 2.5

So......A + B = 0

If we're confining ourselves to real number solutions, Guest is correct !!!

36^2 + b^2  = 15^2

b^2  = 15^2 - 36^2

b^2  = 225 - 1296

b^2  =  -1071

But.....no real number squared can ever be negative......!!!!

The two distances represent legs of a right triangle.....the distance from the ISS  to the communications satellite is the hypotenuse and is given by: :

sqrt [ 249^2 +  378^2 ]  = about   452.6 miles

(2 log(13)+log(26))/(log(196)-log(14))  ≈  3.1784

No prob, rarinstraw......!!!

log(base 3)(12*log(base 16)x=2

log₃ (12*log₁₆ (x) )  = 2      this can be written as .......

3^2  = 12*log₁₆ (x)     

9 = 12* log₁₆ (x )    divide both sides by 12

3/4  = log₁₆  (x)      and we can write

16^(3/4)  = x

[16^(1/4) ]^3  = x

2^3  = x

8 = x

Let's get the lengths of the sides of the triangle, first. Then, we can use the Law of Sines to find C.

A(-1, 3), B(0, 5), and C(4, 3)

BC =.sqrt [ (4 - 0)^2 + (5-3)^2 ]  = sqrtt [16 + 4]  = sqrt(20)

AC = sqrt [ (-1- 4)^2 + ( 3 - 3)^2]=  sqrt [ 25]  = 5

AB =sqrt [ (-1 - 0)^2  + ( 5 -3)^2 ]  = sqrt (5)

So....AC is the hypotenuse.......and we're looking for C....so....using the Law of Sines

sin (90) / AC  = sin C / sqrt(5)

1/5  = sinC / sqrt(5)

sinC = sqrt(5) / 5  =  1 / sqrt(5)

And using the sine inverse

arcsin ( 1 / sqrt(5)   = C  = about 26.57°   = 27°  [rounded]

2^x=576460752303423488

x = log (576460752303423488) / log (2)   =  59

16 - 4 + 8 - 13 + 3 + 4 +1 + 6  + 13   =  34

V_cone =  pi * r^2 * h  / 3

Area of the base  =  pi*r^2  = 70   ...so...

350  = 70 * h / 3  

350  = (70/3) * h     multiply both sides by 3/70

350 (3/70)  = h  = 15 m