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Thanks!

Let's go from here :

110  =  73(1.0235)^x     divide both sides by 73

110/ 73   = (1.0235)^x      take the log of both sides

log (110 / 73)  =  log (1.0235)^x

And, by a log property,   log(a)^x  =   x *log(a)...so.....the right side becomes

log (110 / 73)  =  x * log(1.0235)       divide both sides by log(1.0235)

log (110 / 73)   / log (1.0235)   =  x  ≈  17.65  years

So.....2006 + 17.65   =   2024  [rounded to the closest year ]

Great thanks :-)

If we waned to stretch things.......based on the pattern, the nth term is given by :

1 /  [  (2n+ 1)(-1)ⁿ ]

I was raised Catholic, but it has always amazed me how crazy the Christian religion is. A god tortures his son to death to forgive and save humanity from their disgusting, sinful nature. The day this happened is called Good.  

We are told we should be like god and forgive.

OK. I should torture my dog to death so I can forgive my neighbor for being a disgusting jerk.

Guess what would happen to me if I actually did something like that. My neighbors would crucify me! Kind of ironic, is it? 

Comments anyone.

Thanks :)

Both of these can be solved by manipulating the Law of Cosines  as follows :

arccos [ (a^2 - b^2 - c^2) / [ -2(b)(c)] = A  

Where  "a" is the side opposite the angle we are trying to find [Angle "A"] and "b" and "c" are the other two sides

For the first, we have that

arccos  [ ( 4^2 - 7^2 - 7^2) / [ -2(7)(7)]   = A  ≈  33.2°  = 33°

For the secomd, we have

arccos  [ ( 6^2 - 7^2 - 10^2) / [ -2(7)(10)]   = E  ≈  36.18°   =  36° 

Based on these....see if you can do the last one....if you get stuck.....let us know.....

For these problems, we can use the law of cosines:

c² = a² + b² — 2abcosC

1.

c = JK

a = 3

b = 5

C = 62º

Plug these values into the law of cosines.

(JK)² = 3² + 5² - 2(3)(5)cos 62

(JK)² = 9 + 25 - 30cos 62

(JK)² = 34 - 30cos 62

JK = \(+\sqrt{34 - 30cos 62} \approx4.5\text{ mm}\)

2.

c = EF

a = 6

b = 11

C = 40º

Plug these values into the law of cosines.

(EF)² = 6² + 11² - 2(6)(11)cos 40

(EF)² = 36 + 121 - 132cos 40

(EF)² = 157 - 132cos 40

EF = \(+\sqrt{157 - 132cos 40} \approx7.5\text{ ft}\)

The first says that     4^x  =  3

The second says that  2^(kx)  = 27     

Re-writing the first, we have

 [ 2^2]^x  = 3

[2^x]^2  = 3    cube both sides

( [2^x]^2 )^3   = 27

2^(6x)  = 27

Which implies that  

2^(6x)  =   2^(kx)

Which implies that  k  = 6

Present population=73 million in 2006

Future population =110 million in ??

110,000,000 =73,000,000 x 1.0235^n divide both sides by 73,000,000.

1.506849315 =1.0235^n take the log of both sides

n =Log(1.506849315) / Log(1.0235) 

n =~17.65 years.

However, for "natural growth" in population of people, animals, plants.....etc., it is more accurate to use the natural log(ln, and e);

110 =73 x e^(0.0235n)

1.506849315 =e^(0.0235n) take the natural logs of both sides

n =17.45 years - This is slighly more accurate.

(1,5), (0,6), and (2,3)

We have this form

y = a(x - h)^2  + k       where "a" determines the width (and direction - "up" or "down" ) of the parabola, and (h,k) is the vertex

So  we know that

5  =  a ( 1 - h)^2 + k     →  5 = a(1 -2h + h^2) + k  →  5 = a -2ah + ah^2 + k   (1)

6  = a(0 - h)^2 + k  →  6 = ah^2 + k   (2)

3 = a(2 - h)^2 + k   →  3 = a(4 - 4h + h^2) + k → 3 =  4a -4ah +ah^2 + k     (3) 

Sub ( 2) into  (1)  and (3)

5 =  a - 2ah + 6     →  -1  =  a - 2ah   ( 4)

3  = 4a - 4ah + 6  →  -3 = 4a - 4ah     (5)

Multiply  (4) by -2 and add it to (5)

-1  =  2a    →  a  = -1/2

Using (4)  to find h, we have

-1 = (-1/2) - 2 (-1/2)h

-1/2  =  h

Using (2)  to find k, we have

 6 = (-1/2)(1/4) + k

k = 6 + 1/8  =  49/8

So..........our equation is

y = (-1/2)(x + 1/2)^2 + 49/8

Here's the graph with the points of interest  : https://www.desmos.com/calculator/vnyrl52lp8

wow, nice!

ok stop...........

We can use the tangent here

tan (32)  = H / 68      where H is the building height.......multiply both sides by 68

68 * tan (32)   =  H   ≈  42.5 ft

You're asking about "empirical" probability as opposed to "theoretical" probabilty......they are not necessarily the same.....

For instance......if we flipped a coin 100 times........we should expect that "heads" would occur 50 times  =  50/100 = 1/2  = 50% probability of heads.....this iis the theoretical probability

However......if you actually do this.......the coin may actually come up "heads"  either more or less than 50 times........this is the "empirical" probability (the "observed" probability)

So....1%  =  1/100 chance, theoretically, of something happening......however....the empirical probabilty may be more or less depending upon the observed results

Any non-repeating (or, repeating) decimal can always be expressed as a rational number

1.799  =  1 + 799/1000  =  [1000 + 799] / 1000  =   1799/1000

divide 180 by 3 which is 60

Yours is correct, hectictar.........( target area of triamgle) / (whole area of triangle) = desired probability 

-8k - 33 = -2k + 5(-5k + 1)

Distribute.

-8k - 33 = -2k - 25k + 5

Combine like terms.

-8k - 33 = -27k + 5

Add 27k to both sides.

-8k + 27k - 33 = 5

Add 33 to both sides.

-8k + 27k = 5 + 33

19k = 38

Divide both sides by 19.

k = 38/19

k = 2

p(x)=(x ​2 ​​ +3x+2)(x+1

Do you need the zeros? Or the first derivative?

\(p(x)=(x ^2 ​​ +3x+2)(x+1)=0\\x+1=0\\\color{blue}x_1=-1\)

\(x^2+3x+2=0\\x=-\frac{p}{2}\pm(\sqrt{(\frac{p}{2})^2-q}=-1.5\pm(\sqrt{1.5^2-2}\)

\(x=-1.5\pm0.5\)

\(x_2=-1\)

\(x_3=-2\)

x1 and x2 = -1 is a double zero in a minimum of the function.

\(p(x)=(x ^2 ​​ +3x+2)(x+1)=x^3+3x^2+2x+x^2+3x+2\)

\(p(x)=x^3+4x^2+5x+2\)

\(p'(x)=3x^2+8x+5\)

\(3x^2+8x+5=0\)

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

\(x = {-8 \pm \sqrt{64-60} \over 6}\) 

\( x_{max}=-\frac{5}{3}\\ \\x_{min}=-1\)

  !

IT WAS THE DAY GOD WAS CRUSIFIED

Your cheese aint good your cheese is c.r.a.p 

Yep... dividing is like dealing out a deck of cards to play a game.

If you want to know what 52/4 is, do this:

Deal an entire deck of 52 cards out to 4 players.

The number of cards each player has after you're finished dealing = 52/4