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The side of the square  = the side of the equilateral triangle  = 48/4  = 12 in

And I assume you want to know the perimeter of the shaded figure  ???

There are 6 sides to the shaded figure   each  =  12 in

So...the perimeter  =  6 * 12  =  72 in

The line will be  x  =  the x coordinate of the vertex    =   k

The y coordinate of the vertex  =   - (-3) / 2(-2)  =  3 / -4  =  -3/4

The x coordinate of the vertex  =   -2(-3/4)^2 - 3(-3/4) + 5  =  -2(9/16) + 9/4 + 5  = 

(-9/8) + 18/8 + 40/8    =  49/8 

So....the line is

x  = 49/8  = k

Here's a graph :  https://www.desmos.com/calculator/yiurzdo46y

x^2 + y^2  = 14x + 48y         rearrange as

x^2 - 14x  + y^2 - 48y  = 0      complete the square on x and y

x^2 -14x + 49 + y^2 - 48y + 576  =  49 + 576        factor  and simplify

(x -7)^2  + ( y - 24)^2  =  625

This is a circle centered at  ( 7 , 24)   with  a radius of √625  =  25

The max value for y will therefore be   24 + 25   =  49

The distance between the closest points will occur on the line segment that joins the two centers of the circles

The length of this segment  is

√ [ (17 - 2)^2 + ( 10 - 2)^2 ]   =  √ [ 15^2  + 8^2]  =  √289   =   17 units

From this, we want to subtract the two radii of the circles, i.e., 2  and 10

So....the distance between the closest points  =  17 - 2 - 10  =  17 - 12  =  5 units

Here's a pic : 

If this has two real roots, then  the discriminant must be > 0

So

15^2  - 4(8)c >  0

225 - 32c >  0

225 > 32c

c < 225/32  ⇒ c  <  7.03125

So....the positive integer values for c  that produce two real roots are  [ 1, 7 ]  

And their product  = 7!   =  5040

y = ax^2  + bx  +  c

The x coordinate of the vertex =  -b / (2a)   =  5  ⇒   -b = (2a)*5  ⇒  b = -10a

So  we have this system

a(5)^2  - 10a(5) + c  = 3   ⇒  25a - 50a + c = 3  ⇒  -25a + c  = 3     (1)

a(2)^2 - 10a(2) + c  =  0  ⇒  4a - 20a + c  =  0  ⇒  - 16a  + c  = 0  ⇒  16a - c  = 0   (2)

Add (1)  and (2)    and we get that

-9a  = 3

a  =  -1/3

b = -10 (-1/3) = 10/3

And using  (2)    16 (-1/3) - c  = 0   ⇒  -16/3  = c

So....

y  = (-1/3)x^2  + (10/3)x - 16/3

a  = -1/3      b  =  10/3     c  = -16/3

And their sum is    -7/3

-2x^2 + 4x + 5       factor out the "-2"

-2 ( x^2 - 2x  - 5/2)      

Complete the square on x......take (1/2) of 2  = 1....square iy 1^2  = 1....add and subtract it within the parentheses

-2 ( x^2  - 2x + 1 - 5/2 - 1)     factor the first three terms in the parentheses.....simplify the rest

-2 [ ( x - 1)^2   -  7/2 ]      distribute the "-2"  

-2 ( x - 1)^2 +  7

k = 7

We just need to find the x intersection points  of the parabola and the line y  = 7

So....we have that

7  = 2x^2 + 8x + 4       subtract 7 from both sides

2x^2 + 8x - 3  = 0     

Uing the quadratic formula

x  =  -8 ±√ [ 8^2 - 4(2)(-3) ]       =       -8 ±√ [ 88 ]      =    -8 ± 2√ 22    =     -2 ±√22 /2  =  - 4 ±√22

       __________________             __________         _________                                  ________

                 2 * 2                                      4                          4                                                 2

So....the two points  of  intersection   are  - 4 + √22      and  - 4 - √22

                                                                  _______             _______

                                                                        2                         2

And the square of the distance between these points  =  the area of the square  =

[  (  - 4 + √22) / 2   -  ( -4 - √22) / 2 ] ^2   =

[   2 √22 / 2 ] ^2  =

[ √22] ^2  =

22 units^2

can i still get some cake

ln(4x^2−36x+72)  =        take out the GCF  =  4

ln ( 4 [ x^2 - 9x + 18 ] ) =      factor  again

ln [ 4 ( x - 3) ( x - 6 ) ] =   

Use the property that.....  ln (a * b * c)  =  ln a + ln b + ln c

ln 4  +  ln ( x - 3)  +  ln ( x - 6)

The thing to expand is this 

ln(4x^2−36x+72)

Here's my attepmt at part "b"

To the nearst cm.....the ramp length  is 120cm

So....the angle,  θ, in this case  can be found as

sin   θ =    27.43  / 120

To find   θ  use the arcsin  function

arcsin ( 27.43 / 120 )  =   θ ≈  13.21°

Percentage error  =    [ 13.21  - 13 ]  / 13    ≈ .016  =  1.6%

Here's the "a" part

sin (13°)  =  27 /  Ramp Length            rearrange as

Ramp Length  =    27  /  sin (13°)   ≈   120.03  cm

Here : https://web2.0calc.com/questions/help-help-quick-quick

The domain is only affected by the values of x that make the denominator 0

So....set   x^2 + 4x - 21  = 0  to find these values

Factor as

(x + 7) ( x - 3)  = 0        set each factor to 0   and solve for x   and we have

x + 7  = 0                   x - 3  = 0

x  = - 7                        x  =  3

So  the domain is all x except  x  = -7  and x =  3

In interval notation

( - infinity, -7)   U  (-7, 3)   U   ( 3, infinity )

Thanks a lot!! 

Since we have vertical asymptotes at  x  = 1  and x  = - 2, the denominator has the form

(x - 1) (x + 2)    since  x = 1  and x  = -2   will make the denominator 0

So....expanding....we  have   x^2 -  1x  + 2x  - 2  =  x^2 + 1x  - 2 = x^2 + ax + b

So

a  =1     and b   =-2

And  a + b  =  1 + (-2)  =   -1

Thank you very much! 

We need to know the radius of the Earth (in meters)   ≈ 6,370,000 m

We have a right triangle with a hypotenuse of the radius of the Earth + height of lighthouse above sea level =  [6,370,000 + 140 + 15 ] m = 6,370,155 m

The distance from the center of the Earth to the ship  is just the Earth's radius and forms one leg of the triangle

And the distance we are looking for is the other leg of the triangle  = D

So....by the Pythagorean Theorem :

D  = √ [ 6,370,155^2 - 6,370,000^2 ]  ≈  44,438 m  =  44.438 km  ≈ 27.5 miles

Very nice, heureka !!!!

Thank you, GingerAle !

if line 3 is not wrong, why don't you get  $\cos 1^{\circ} = \cos 1^{\circ}$ ?

I got it out !

What did I do wrong?

$2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} = 90\cdot \cot 1^{\circ}$

$\small{ \begin{array}{|rcll|} \hline \mathbf{2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 176 \sin 176^{\circ} + 178 \sin 178^{\circ} + 180 \sin 180^{\circ} } &=& \mathbf{90\cdot \cot 1^{\circ}} \\\\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin 92^{\circ} + \cdots + 176 \sin 176^{\circ}+178 \sin 178^{\circ} + 180 \underbrace{\sin 180^{\circ}}_{=0} &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin 92^{\circ} + \cdots + 176 \sin 176^{\circ}+ 178 \sin 178^{\circ} &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin (180^\circ -92^{\circ}) + \cdots + 176 \sin (180^\circ-176^{\circ} ) + 178 \sin (180^\circ-178^{\circ} ) &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} + 92 \sin 88^{\circ} + \cdots + 176 \sin 4^{\circ}+178 \sin 2^{\circ} &=& 90\cdot \cot 1^{\circ} \\ 2 \sin 2^{\circ} + 4 \sin 4^{\circ} + \cdots + 88 \sin 88^{\circ}+ 90 \sin 90^{\circ} \\ 178 \sin 2^{\circ} + 176 \sin 4^{\circ} + \cdots + 92 \sin 88^{\circ} &=& 90\cdot \cot 1^{\circ} \\ (2+178)\sin 2^{\circ}+(4+176)\sin 4^{\circ} + \cdots +(88+92)\sin 88^{\circ}+ 90 \sin 90^{\circ} &=& 90\cdot \cot 1^{\circ} \\ 180\sin 2^{\circ}+180\sin 4^{\circ} + \cdots +180\sin 88^{\circ}+ 90 \sin 90^{\circ} &=& 90\cdot \cot 1^{\circ} \quad | \quad : 90 \\ 2\sin 2^{\circ}+2\sin 4^{\circ} + \cdots +2\sin 88^{\circ}+ \sin 90^{\circ} &=& \cot 1^{\circ} \\ 2\sin 2^{\circ}+2\sin 4^{\circ} + \cdots +2\sin 88^{\circ}+ \sin 90^{\circ} &=& \dfrac{\cos 1^{\circ}}{\sin 1^{\circ}} \quad | \quad \cdot \sin 1^{\circ} \\ 2\sin 1^{\circ}\sin 2^{\circ}+2\sin 1^{\circ}\sin 4^{\circ} + \cdots +2\sin 1^{\circ}\sin 88^{\circ}+ \sin 1^{\circ}\sin 90^{\circ} &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 3^{\circ}+ \cos 3^{\circ}-\cos 5^{\circ} + \cdots + \cos 87^{\circ}-\cos 89^{\circ}+ \sin 1^{\circ}\sin 90^{\circ} &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 89^{\circ}+ \underbrace{ \sin 1^{\circ}\sin 90^{\circ}}_{=\frac12 (\cos 89^\circ - \cos 91^\circ )} &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 89^{\circ}+ \frac12 (\cos 89^\circ - \cos 91^\circ ) &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\cos 89^{\circ}+ \frac12 \cos 89^\circ -\frac12 \cos 91^\circ &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\frac12 \cos 89^\circ -\frac12 \cos 91^\circ &=& \cos 1^{\circ} \\ \cos 1^{\circ}-\frac12 (\cos 89^\circ + \cos 91^\circ) &=& \cos 1^{\circ} \\ \boxed{ \cos 91^\circ = -\cos(180^\circ -91^\circ) \\ = -\cos 89^\circ }\\ \cos 1^{\circ}-\frac12 (\cos 89^\circ -\cos 89^\circ ) &=& \cos 1^{\circ} \\ \cos 1^{\circ}-0 &=& \cos 1^{\circ} \\ \mathbf{ \cos 1^{\circ} } &=& \mathbf{ \cos 1^{\circ} } \\ \hline \end{array} }$