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Note that we have a right triangle..... and  since they are co-functions...sin P  =  cosR

And cos P  =  sin R

So

sin R  - cos R  =

cos P  - sin P   =

0.8192   -  0. 5736  =

0.2456

1.    Correct   !!!!

2.   Not sure about this...but... I believe that you are correct

3.   Correct !!

4. The radius of circle V is 16.  Correct!!

Also...Circles V  and W  are similar because all circles are similar...

5. Correct  !!!

\(7j^4c^4\sqrt{2j}\)

Square and square root the expression.

\(=\,\sqrt{\bigg(7j^4c^4\sqrt{2j}\bigg)^2}\\~\\ =\,\sqrt{7^2(j^4)^2(c^4)^2(\sqrt{2j})^2\vphantom{\bigg)}}\\~\\ =\,\sqrt{49j^8c^8(2j)\vphantom{\bigg)} }\\~\\ =\,\sqrt{49\cdot2\cdot j^8\cdot j\cdot c^8\vphantom{\bigg)} }\\~\\ =\,\sqrt{98j^9c^8 \vphantom{\big)}}\)

Call the height, h

Since the apex angle  is a right angle... the height is forms a geometric mean, thusly 

16/h  = h/4      cross-multiply

h^2  =  64

h  =  8

And by the Pythagoren Theorem

√ [ h^2  + 4^2  ]  =  y

√ [ 8^2 + 4^2 ] = y

√80  =  y

√[16 * 5]   y

4√5  = y

y / 10   =  [  y + 4  ]   / 15   cross-multiply

15y  = 10y + 40

5y  =  40

y  = 8  

Let's call the unlabeled side of the large triangle  " x ".

And let's call the height of that triangle  " h ". 

By Pythagorean theorem, we can make the equation

x² + y²   =   (16 + 4)²

                                    Subtract  y²  from both sides of the equation.

x²   =   (16 + 4)² - y²

Again by the Pythagorean theorem, we can make the equation

16² + h²   =   x²

                            Subtract  16²  from both sides of the equation.

h²   =   x² - 16²

                                             Substitute   (16 + 4)² - y²   in for   x²

h²   =   (16 + 4)² - y² - 16²

Again by the Pythagorean theorem, we can make the equation

4² + h²   =   y²               Substitute   (16 + 4)² - y² - 16²   in for  h²

4² + (16 + 4)² - y² - 16²   =   y²      Now we can solve for  y .

                                                     Simplify the left side of the equation.

4² + 20² - y² - 16²   =   y²

16 + 400 - y² - 256   =   y²

160 - y²  =  y²

                          Add  y²  to both sides of the equation.

160   =   2y²

                          Divide both sides by  2

80   =   y²

                          Take the positive square root of both sides.

√80  =  y

y  =  √[2*2 * 2*2 * 5]

y  =  4√5

( ⁿ√a) ^m   can be written as    a^m/n  .....so.....

⁵(√a⁵ )  =    a^5/5  =  (a^1/5)⁵  = a  ⇒   Answer  C

Second one

³√ [216a³b9 ]  =  ³√216  * ³√a³  *  ³√b⁹    =  (6³)^1/3 * (a³)^1/3 * (b⁹)^1/3  = 

6ab³ ⇒ Answer C

Last one....look at answer B

√[98 j⁹c⁸]  = √{49*2]  * √ [j⁹ ] * √[c⁸ ]   =

√49  * √2  * √ [j⁸]  * √j  * √[c⁸ ]   =

7  * √2  *  (j⁸)^1/2  * √j  *  (c⁸)^1/2    =

7j⁴c⁴√[ 2j ]   

log ( x + 2)  = 1 + log (x)

We can write

log ( x + 2)  -  log (x)  =  1       using a log property, we have that

log [ (x + 2) / x ]   = 1       in exponential form, we have

10 ^1   =  [x + 2] / x

10  =  [ x + 2] / x      multiply both sides by x

10x  = x + 2   subtract x from  both sides

9x  =  2    divide both sides by 9

x  = 2/9  ≈ 0.222

That was wrong, the answer was 11, but thank you!

Thanks everyone!

15/10 = 1.5 

so y= 4/1.5

y= 2.666666667 

A regular hexagon has perimeter p and area A. Compute (p^2)/A.

30  out of 40 dislike the new start time  = 3/4

So....we would expect that   (3/4)*1000  =  750  to dislike it

Call a side,  S

The perimeter  is 6S

The area  = 6* (1/2)S^2 (√3) /2   = 6*(√3/4) *S^2  = (3√3/2)S^2 

So  

P^2  / A  =    (6S)^2  / [ (√3/3/2) *S^2 ]   =  

  36                        72               24

______   =            ___    =       ___    =  8√3

(3√3) /2                3√3              √3

Can You help me with this question too

We can use a weighted average here  =

[ sum of  freq* mipoint ] / [sum of freq ]  \

[  (1*5) + (4*15)  + (7 * 25)  + (2 * 35)  + (3 * 45)  + (3 * 55)

_______________________________________________

     [ 1 + 4 + 7 + 2 + 3 + 3 ]

[  610 ]

______

   20

30.5  cm  =  mean height of the plant

Area  of a trapezoid  =

height  * ( sum of base lengths)  / 2

So...let the height  = h

One  base  = h + 3

And the other is h + 5

So...we have this

375  =  h [  (h + 3  + h + 5 ]  / 2      multiply both sides by 2

750  =  h [ 2h + 8 ]

750  = 2h^2 + 8h      rearrange as

2h^2 + 8h   -  750   =  0 

Using the quadratic formula....we get the answer  ≈  17.468  =  17.5 yards (rounded)   

"b" is correct

This is pretty simple....believe it, or not    !!!

1.   x^2  +  12x

Take  1/2  of 12  =  6

Square this  = 6^2  =  36.....and that's what we need to add to complete the square  !!!!

2. x^2 - 8x

Take (1/2)  of -8  = -4

Square this  = (-4)^2  =   16

3. x^2  - 18x

Take (1/2) of -18  = -9

Square this  (-9)^2   =  81

log (2x)  + log (x -5)    = 2

We can use this rule to simplify things, Manuel

log a  + log b  =  log (a * b)....so ....we can write the left sides as

log[ (2x * (x -5) ]  = 2

log [  2x^2 - 10x]  =  2      in exponential form we have

10^2  = 2x^2 - 10x

100  =  2x^2  - 10x        rearrange as

2x^2 - 10x - 100  = 0       divide through by 2

x^2 - 5x -50  =  0        factor as

(x - 10)(x + 5)  = 0

Setting each factor to 0 and solve for x  and we have that

x  = 10     or   x  = -5

We must reject  x  = -5   because it would mean that we would be taking logs of negative numbers in the original equation

So.....x  = 10

Thank you so much. I have been stuck on this one.

Let x  be the smaller dimension of the window.....then 1.2 x  is the other dimension

So we have

 x * (1.2x)  = 811.2

1.2x^2  = 811.2       divide both sides by 1.2

x^2  = 676       take the square root of both sides

x  = 26 in    this is the smaller dimension

1.2 (26)  =  31.2 in      this is the larger dimension

So.....  "a"  is correct

None of the answers are fully factored

"B"   is correct...however......(3x - 6y)(3x - 6y)   can be factored further as

3(x - 2y) * 3 ( x - 2y)  =  9 (x -2y)² 

9x^2 -36xy + 36y^2

Take out the greatest common factor

9 (x^2 - 4xy + 4y^2)

Factor the polynomial in theparenthese

9 (x - 2y) (x - 2y)

9 (x - 2y)²

my pleasure 

Im doing good thanks for asking