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Yep, that's correct. Thank you!

1)

A 30-60-90 triangle has a specific ratio in which its side lengths are defined. It is a \(1:\sqrt{3}:2\) ratio. Knowing this set ratio, it is easy to determine that the hypotenuse is double the length of the shorter leg. Therefore, the hypotenuse length is 16 meters long!

2) 

This question is very similar to the previous triangle, except that we are dealing with a different special right triangle. A 45-45-90 triangle's ratio of the side lengths is \(1:1:\sqrt{2}\). If one of the lengths of the legs is 8 inches, then the hypotenuse has length \(8\sqrt{2}\text{ inches}\).

3) 

It appears as if this question has a formatting issue. 

divide 9(cos5pi/4+i sin5pi/4)/3(cos pi+i sin pi)

\(9(cos5pi/4+i sin5pi/4)/3(cos pi+i sin pi)\\ =\frac{9e^{(5\pi/4)i}}{3e^{(\pi/4)i}}\\ =\frac{3e^{(5\pi/4)i}}{e^{(\pi/4)i}}\\ =3e^{(5\pi/4)i-(\pi/4)i}\\ =3e^{(4\pi/4)i}\\ =3e^{\pi i}\\ =3(cos\pi+i sin\pi)\\ =3(-1+0)\\ =-3\)

C'mon, Manuel....you have posted a lot of questions JUST like this one....just use the calculator on this website..

it is easy.....

Enter    2

Enter    6

Push the LOG key

Push the  =   key........    

Thank You!!!!!

I assume that we have  log (y / x)

We are using the property

Log (a / b)  =  log a  - log b

So   

log (y / x)  =  log y - log x

...never mind.....  Chris has done it....

See here  :

https://www.desmos.com/calculator/valyot2vsy

Sorry....got that wrong...will try again...

log x^3  =   3 log x    according to the rules of logs

f (f^-1 (x) )  = x        = 2010

Urm, the second and third statements are correct?

At year ZERO    x = 0   

Substituting this into the equation yields

f(x) = 500/20 = 25  employees

At year 2  (x=2)   yields   74  

At year 3   (x=3)  yields   employees = 120

After sa certain number of years the  e^-0.6x becomes very very small  (~0) and the maximum employees (500) will be reached.

Can you pick out the true statements now?

2)

Here is my attempt at this:

E Q U A L S

Since the first and the last letters are fixed at L and Q respectively, then the remaining 2 letters out of 4 can be permuted in 4P2 =12 ways as follows:

LAEQ, LASQ, LAUQ, LEAQ, LESQ, LEUQ, LSAQ, LSEQ, LSUQ, LUAQ, LUEQ, LUSQ (12)

Thank you

The last graph?

....so the antilog of 9     would be    10^9 = 1 000 000 000  

Slope = rise/ run  or   (y1-y2) / (x1-x2)

for the first two points

slope = (4-3) / (2-6) = - 1/4

for the second and third points we need slope = -1/4 , then solve for c

(3-c) / (6-(-5) ) = -1/4

3-c = -1/4(11)

3-c = -11/4

3+11/4 = c

23/4 = c     or  5 3/4   or  5.75

a²  = k / b³

7²  = k / 3³

49  =  k / 27     ⇒  k   =  49 / 27

When b  = 6

a²  =  49 / (27*6³ )

a²  =  49 / 5832

Hmmmm.... that looks pretty simplified already.... but you COULD do this

according to the rules of logs

Log (xy) = log(x) + log (y)

Note the series  is

3, 7, 11, 15.....

So .....the recursive rule is

a_n  = a_n-1 + 4, a₁  = 3  ⇒     first answer

f(1)  = 2  ⇒  f^-1(2)  = 1  

f(2)  =  6     ⇒  f^-1(6)  = 2

So

f^-1 (f^-1 (6) )   =   f^-1 (2)  =   1

9 is an exponent  on the base  - (usually base 10, if not specified)

Thanks so much!

Note that 

C(n, 0) + C(n+1,1) + C(n + 2, 2) + C(n + 3, 3) +.....+ C(n + m, m)  = C(n + m + 1, m)

So

C(4,0)  + C(5, 1) + C(6,2) + C(7,3)  = C(8,3)   = 56