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-100 * 2^(0.5y)  = -5   divide both sides by -100

2^(0.5y)  =  -5/-100  = 1/20    take the log of both sides

log 2^(0.5y)  = log (1/20)      and we can write

(0.5y) * log 2  = log (1/20)

y  =  log(1/20) / (0.5* log 2 ]

y = (1/0.5) *  log (1/20) / log 2         reverse the change-of-base theorem

y =  2  * log₂  (1/20)

1.  The common difference  = 6

Recursive rule

a_n =  a_n-1 - 6,  a₁  = 12

2.      Here : https://www.desmos.com/calculator/2bfhyo27zs

√[7] x (√x − 7√7)  =

x√[7x]  -  √[7] x * 7√7  =

x √[7x]  -  7x √ [ 7 * 7 ]   =

x √[7x]  -  7x √ 49  =

x √[7x]  - 7*7x  =

x√[7x]  - 49x   =

x [ √[7x] - 49  ]

1. First five terms  =  10, 8.5, 7, 5.5, 4

We have a constant difference between terms....so...arithmetic

2. Already covered

3.I think this is what is wanted :   https://www.desmos.com/calculator/0ajkqjacqu

Just  7

Is all of "7x"  under the first radical, or just the " 7?"

1. Explain how you know if a radical expression is in simplest form

If you have completely taken all perfect squares, cubes, etc., out of the radical.....it is in simplest form

2. How can you tell if radicals are like radicals?

Simplify both......if both are the same.....they are like radicals

3. Is the inverse of a function always a function? Explain.

If the original function isn't one-to-one, then the inverse isn't a function

Thank you!! 

THX, Melody.....I don't like the fact that perfectly good questions (and their answers) are often  "locked"

arcsin (0.3)  = 17.45° =  t

arccos(0.48)  =  61.31°  =  t

Looks like this one is False

Another way we can prove this

1 - sin^2 t  =  cos^2 t

1 - (0.3)^2  = (0.48)^2   ???

1 - .09  =  .2304  ????

.91  = .2304   ???......False

Thank you so much, CPhill! 

Remember that as cos(theta)  goes from 0 to 90  degrees....its value decreases from 1 to 0

So.....

t  must be an angle  <  w

So....

Angle w  must be >  Angle t

The second one is ±0.11

Thank you! 

From his line of sight....we have that

tan (20)  =  height of street lamp from his line of sight / distance from streetlamp

tan (20)  =  height of streetlamp from his line of sight  /  25    ..... multiply both sides by 25

25* tan (20)  =  height of streetlamp from his line of sight

So......the total height of the streetlamp  = 

6 + 25 * tan(20) ≈  15 ft

Let the vertex be at (0,0)

And the point (12, 4)  is on the parabola.....so we have the form

y  = mx^2     and we know that

4  = m(12)^2

4  = 144m      divide both sides by 144

1/36  =  m

So.....the  form is

y = (1/36)x^2    

36y  = x^2

And   36  = 4a      divide both sides by  4

9  = a

So...the focus  is given by  (0, 0 +a)  = (0, 0 + 9)  =  (0, 9)

So.....the microphone should be placed 9 inches from the vertex

Thank you! 

sec  is positive in quads 1  and 4

tan is negative in quads 2 and 4

So....the intersection of these is quad 4   =  270° < x < 360°

Arc length  =   r * (theta in radians)

35  =  15 * (theta in radians)      divide both sides by 15

35/15  =  theta in radians  =  7/ 3  ≈  2.33.... rads

Also.....a 6-8-10   triangle is right  [ It is a multiple of the basic 3-4-5 right triangle ]

Circumference of ENTIRE circle is  pi * diameter =  30 pi   inches

Entire circle is 2 pi Radians

this arc is a fraction of 2 pi radians     the fraction  is  35 in / 30 pi   inches

35/(30pi) * 2pi Radians = 2.33 Radians

The total circumference is  pi x diam        = pi x 20(2) = 40 pi   ft

38 degrees is 38/360 fraction of this total circumference     38/360 x  40 pi = 13.26 ft

We actually know two legs of a right triangle.....

We need to find the hypotenuse  = √[50^2 + 75^2 ] 

So....the cosine of the angle of elevation  =   75 / √[ 75^2 + 50^2 ]  ≈   0.83 

pi/7 Radians   x  180 degrees/ pi Radians = 25.71 degrees