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Verify the following identity:
sin((3 x)/2) (sin(x))/(sin(x/2)) = sin(x) + sin(2 x)

Multiply both sides by sin(x/2):
sin(x) sin((3 x)/2) = ^?sin(x/2) (sin(x) + sin(2 x))

sin(x) sin((3 x)/2) = 1/2 (cos(x - (3 x)/2) - cos(x + (3 x)/2)) = 1/2 (cos(-x/2) - cos((5 x)/2)):
(cos(-x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) (sin(x) + sin(2 x))

Use the identity cos(-x/2) = cos(x/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) (sin(x) + sin(2 x))

sin(x/2) (sin(x) + sin(2 x)) = sin(x/2) sin(x) + sin(x/2) sin(2 x):
(cos(x/2) - cos((5 x)/2))/(2) = ^?sin(x/2) sin(x) + sin(x/2) sin(2 x)

sin(x/2) sin(x) = 1/2 (cos(x/2 - x) - cos(x/2 + x)) = 1/2 (cos(-x/2) - cos((3 x)/2)):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(-x/2) - cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

Use the identity cos(-x/2) = cos(x/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2) - cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

(cos(x/2) - cos((3 x)/2))/(2) = 1/2 cos(x/2) - 1/2 cos((3 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + sin(x/2) sin(2 x)

sin(x/2) sin(2 x) = 1/2 (cos(x/2 - 2 x) - cos(x/2 + 2 x)) = 1/2 (cos(-(3 x)/2) - cos((5 x)/2)):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos(-(3 x)/2) - cos((5 x)/2))/(2)

Use the identity cos(-(3 x)/2) = cos((3 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2) - cos((5 x)/2))/(2)

(cos((3 x)/2) - cos((5 x)/2))/(2) = 1/2 cos((3 x)/2) - 1/2 cos((5 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2))/(2) - (cos((5 x)/2))/(2)

(cos(x/2))/(2) - (cos((3 x)/2))/(2) + (cos((3 x)/2))/(2) - (cos((5 x)/2))/(2) = 1/2 cos(x/2) - 1/2 cos((5 x)/2):
(cos(x/2) - cos((5 x)/2))/(2) = ^?(cos(x/2))/(2) - (cos((5 x)/2))/(2)

1/2 (cos(x/2) - cos((5 x)/2)) = 1/2 cos(x/2) - 1/2 cos((5 x)/2):
1/2 cos(x/2) - 1/2 cos((5 x)/2) = ^?1/2 cos(x/2) - 1/2 cos((5 x)/2)

The left hand side and right hand side are identical:
Answer: | (identity has been verified)

You should answer your own goddammn question. The answer is the sum of all your meters. You are in the habit of having someone do all your work for you.  You will not learn the material, or how to think on your own, if you choose this path

I'll be happy to do math that I am assigned! Yay!

To figure this out, we must know the base comparisons. This just requires memorization, unfortunately.

\(1in=0.0254m\)

I happen to know that this is the base comparison. However, I'll have to do 2 conversions. One for the appropriate number of inches and one for meters to millimeters. Here it goes!

Bam! Done! 12 inches = 304.8 millimeters!

Point X is on overline AC such that AX = 3 CX = 12. If angle ABC = angle BXA = 90 degrees, then what is BX?

\(\begin{array}{|rcll|} \hline \overline{BX}^2 &=& \overline{AX}\cdot \overline{CX} \\ \overline{BX}^2 &=& 3\cdot 12 \\ \overline{BX}^2 &=& 36 \\ \mathbf{\overline{BX}} &\mathbf{=}& \mathbf{ 6 } \\ \hline \end{array}\)

Right \(\triangle{ABC}\) has AB = 3, BC = 4, and AC = 5.
Square XYZW is inscribed in triangle ABC with X and Y on \(\overline{AC}\),
W on \(\overline{AB}\), and
Z on \(\overline{BC}\).
What is the side length of the square?

Let s is the side length oft the square \( = \overline{XY} = \overline{YZ} = \overline{ZW} = \overline{WX}\)

Let h = \(\overline{BT}\)

Let A the area of \(\triangle{ABC}\)

h = ?

\(\begin{array}{|rcll|} \hline A &=& \frac{\overline{AB} \cdot \overline{BC} }{2} \\ A &=& \frac{3\cdot 4}{2} \\ \mathbf{A} &\mathbf{=}& \mathbf{6} \\\\ A &=& \frac{\overline{AC}\cdot h}{2} \\ A &=& \frac{5\cdot h}{2} \quad & | \quad \mathbf{A=6} \\ 6 &=& \frac{5\cdot h}{2} \\ \mathbf{h} &\mathbf{=}& \mathbf{ \frac{12}{5} } \\ \hline \end{array}\)

\(\mathbf{\overline{BW} =\ ?}\)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BW} } {s} &=& \frac{ \overline{AB} } { \overline{AC} } \\ \frac{ \overline{BW} } {s} &=& \frac{ 3 } { 5 } \\ \mathbf{ \overline{BW} } & \mathbf{=} & \mathbf{ \frac{3}{5}s } \\ \hline \end{array}\)

\(\mathbf{\overline{BZ} =\ ?} \)

\(\begin{array}{|rcll|} \hline \frac{ \overline{BZ} } {s} &=& \frac{ \overline{BC} } { \overline{AC} } \\ \frac{ \overline{BZ} } {s} &=& \frac{ 4 } { 5 } \\ \mathbf{\overline{BZ}} &\mathbf{=}& \mathbf{\frac{4}{5}s } \\ \hline \end{array}\)

s = ?

\(\begin{array}{|rcll|} \hline A_{\triangle{ZBW}} = \frac{ \overline{BW}\cdot \overline{BZ} } {2} &=& \frac{s\cdot(h-s)} {2} \\ \overline{BW}\cdot \overline{BZ} &=& s\cdot(h-s) \quad & | \quad \mathbf{ \overline{BW} =\frac{3}{5}s } \quad \mathbf{ \overline{BZ} =\frac{4}{5}s } \quad \mathbf{h=\frac{12}{5}} \\ \frac{3}{5}s \cdot \frac{4}{5}s &=& s\cdot(\frac{12}{5}-s) \\ \frac{12}{25}s &=& \frac{12}{5}-s \\ s+\frac{12}{25}s &=& \frac{12}{5} \\ s \cdot \left(1+\frac{12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{25+12}{25} \right) &=& \frac{12}{5} \\ s \cdot \left(\frac{37}{25} \right) &=& \frac{12}{5} \\ s &=& \frac{25}{37} \cdot \frac{12}{5} \\ s &=& \frac{5}{37} \cdot 12 \\ s &=& \frac{60}{37} \\ \mathbf{s} &\mathbf{=}& \mathbf{1.\overline{621}} \\ \hline \end{array}\)

The side length oft the square is \(\mathbf{1.\overline{621}}\)

The three points (3,-5), (-a + 2, 3), and (2a+3,2) lie on the same line.

What is a?

Intercept theorem:

\(\begin{array}{|rcll|} \hline \dfrac{(2a-3)-(3)}{ 2-(-5) } &=& \dfrac{ (-a+2) - 3 } { 3-(-5) } \\\\ \dfrac{2a}{ 7 } &=& \dfrac{ -a -1 } { 8 } \\\\ 16a &=& -7a-7 \\ 23a &=& -7 \\ \mathbf{a} &\mathbf{=}& \mathbf{ -\frac{7}{23} } \\ \hline \end{array} \)

Let triangle ABC be an isosceles triangle such that BC = 30 and AB = AC.
We have that I is the incenter of triangle ABC, and IC = 18.
What is the length of the inradius of the triangle?

\(\begin{array}{|rcll|} \hline \left( \frac{30}{2} \right)^2 + r^2 &=& 18^2 \\ 15^2 + r^2 &=& 18^2 \\ r^2 &=& 18^2-15^2 \\ r^2 &=& 324-225 \\ r^2 &=& 99 \\ r^2 &=& 9\cdot 11 \\ r &=& 3\cdot \sqrt{11} \\ \hline \end{array}\)

r = 9.95

please leave a like because u r flipping fast 

i posted that only a couple min. ago

difference between length and width   =   length - width

                                                             =   272.5 feet - 151.25 feet

                                                             =   121.25 feet

So, the parking lot is  121.25 feet  longer than it is wide.

    3 4

+  2 3

    5 7

Let's call the base " 2s " and then draw in a height to that base.

Look at the triangle that the line for the height makes. It is a  " 45-45-90 "  triangle.

So...

side across from the 90° angle = √2 * (side across from 45° angle)

                                              s  =  √2 * (height)

                                       s / √2  =  height

Now we know the base and the height, and..

area of parallelogram  =  base * height

8√2  =  2s * s / √2                                   Multiply both sides by  √2  .

8√2 * √2  =   2s * s

8 * 2  =  2s²                                            Divide both sides by  2  .

8  =  s²

s  =  ±√8         Since a negative  s  causes a negative side length, just take the positive option.

s =  2√2

What are the integer  x  values such that  y  is bigger than  x  ?

When  x  is  0  ,  y  is about  4

When  x  is  1  ,  y  is about  -0.5

When  x  is  2  ,  y  is about  1

When  x  is  3  ,  y  is about  4

When  x  is  4  ,  y  is about  7

When  x  is  5  ,  y  is about  7.5

When  x  is  6  ,  y  is about  7

When  x  is  7  ,  y  is about  6.5

When  x  is  8  ,  y  is about  10

Also, notice how if you drew the line  y  =  x  , all the x values that have a y value that goes above that line will meet the criteria that y > x  . So....the integer  x  values such that  y  is bigger than  x are:  0, 3, 4, 5, 6, 8

0 + 3 + 4 + 5 + 6 + 8  =  26

96% you motherf*cking @$$hole

Since all the points are on the same line, the slope between each point will be the same.

slope  =  \(\frac{\text{change in y}}{\text{change in x}}\)

 slope between first and second points  =  \(\frac{(-5)-(3)}{(3)-(-a+2)}=\frac{-8}{1+a} \)

slope between second and third points  =  \(\frac{(3)-(2)}{(-a+2)-(2a+3)}=\frac{1}{-3a-1} \)

      slope between third and first points  =  \(\frac{(2)-(-5)}{(2a+3)-(3)}=\frac{7}{2a}\)

Let's pick any two and equate them.

\(\frac{7}{2a}=\frac{-8}{1+a} \)         Cross - multiply...

(7)(1+a) = (-8)(2a)

7 + 7a  =  -16a

7  =  -23a

3x(2x - 1)

=  (3x)(2x) + (3x)(-1)

=  (3)(2)(x)(x) + (3)(-1)(x)

=  6x² - 3x

If the line's slope is  3  and y-intercept is  1 , the slope-intercept form of the line is:

y  =  3x + 1

And the equation for the circle is:

x² + y²  =  1                        We want to find what x is when y = 3x + 1. So substitute 3x + 1 in for y.

x² + (3x + 1)²  =  1

x² + (3x +1)(3x + 1)  =  1

x² + 9x² + 6x + 1  =  1

10x² + 6x  =  0                   Factor out an  x  from both terms.

x(10x + 6)  =  0                   Set each factor equal to zero and solve for  x .

x  =  0       or       x  =  -3/5

Now we can plug these  x  values into the equation of the line to find the  y  coordinate of the intersection points.

When  x  =  0 ,                       When  x  =  -3/5 ,

y  =  3(0) + 1                           y  =  3(-3/5) + 1

y  =  1                                     y  =  -4/5

So...the coordinates of the two points are:    (0, 1)   and   (-3/5, -4/5)

Can be done as follows:

.Need to add a constant of course.