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oh ok

\( 60 = (15\times10 \times sin \theta ) / 2\\ 120 = 150 \times sin \theta \\ \frac{12}{15} = sin \theta \\ 0.8= sin \theta \\ sin\theta = 0.8\\ first\;or \; second \;quadrant\\\)

asin(0.8) = 53.130102354156

\(\theta=(180n + (-1)^n*53)^\circ \qquad n\in Z\)

The point of the question is to find the angle. 

60 = (15 x 10 x sin) / 2

This question does not make sense.

You need and angle after sin ..    :)

T= 36 $      S= 51 $

easy!

4t + 5s = 399

s= t +15

and so on...

Given : sin\alpha+sin\beta=1 , cos\alpha+cos\beta=1

Find the value of sin\alpha-cos\beta

Given:

\(sin\alpha+sin\beta=1,\ cos\alpha+cos\beta=1\)

Find the value of

\(sin\alpha-cos\beta\)

\(\begin{array}{|lrcll|} \hline (1) & \sin(\alpha) + \sin(\beta) &=& 1 \\ (2) & \cos(\alpha) + \cos(\beta) &=& 1 \\ \hline (1)-(2): & \sin(\alpha) + \sin(\beta) - \Big(\cos(\alpha) + \cos(\beta) \Big) &=& 0 \\ & \sin(\alpha) + \sin(\beta) -\cos(\alpha) - \cos(\beta) &=& 0 \\ & \underbrace{\sin(\alpha) -\cos(\alpha)}_{=\sqrt{2}\sin(\alpha-45^{\circ})} &=& \underbrace{\cos(\beta) - \sin(\beta)}_{=-\sqrt{2}\sin(\beta-45^{\circ})} \\ & \sqrt{2}\sin(\alpha-45^{\circ}) &=& -\sqrt{2}\sin(\beta-45^{\circ}) \\ & \sin(\alpha-45^{\circ}) &=& - \sin(\beta-45^{\circ}) \\ & \sin(\alpha-45^{\circ}) &=& \sin\Big(-(\beta-45^{\circ})\Big) \\ & \sin(\alpha-45^{\circ}) &=& \sin(45^{\circ}-\beta ) \\ & \alpha-45^{\circ} &=& 45^{\circ}-\beta \\ & \alpha &=& 90^{\circ}-\beta \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \sin(\alpha) - \cos(\beta) \\ & =& \sin(90^{\circ}-\beta) - \cos(\beta) \\ & =& \cos(\beta) - \cos(\beta) \\ & =& 0 \\ \hline \end{array}\)

Proof:

\(\begin{array}{|rcll|} \hline \sin(\alpha) + \sin(\beta) &=& 1 \quad | \quad \alpha = 90^{\circ}-\beta \\ \sin(90^{\circ}-\beta) + \sin(\beta) &=& 1 \\ \cos(\beta) + \sin(\beta) &=& 1 \quad | \quad \cos(\beta) + \sin(\beta)=\sqrt{2}\sin(\beta+45^{\circ}) \\ \sqrt{2}\sin(\beta+45^{\circ}) &=& 1 \\ \sin(\beta+45^{\circ}) &=& \frac{1}{\sqrt{2}} \\ \beta+45^{\circ} &=& \arcsin( \frac{1}{\sqrt{2}} ) + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \beta+45^{\circ} &=& 45^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \beta &=& 90^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\\\ \alpha &=& 90^{\circ}-\beta \\ \alpha &=& 90^{\circ}-(90^{\circ} + n\cdot 360^{\circ}) \qquad n \in \mathbb{Z} \\ \alpha &=& 0^{\circ} + n\cdot 360^{\circ} \qquad n \in \mathbb{Z} \\ \hline \end{array}\)

5^3 = 5 * 5 * 5 = 125      

−6+log(m+7)=−3

\(log(m+7)=3\\ 10^{log(m+7)}=10^3\\ m+7=1000\\ m=993 \)

(x-5)(x+2)(2x-1)=200

Your question makes no sense.

Let the cost of a tie  = T

Then.....the cost of a shirt  = T + 15

And we have that

4T + 5(T + 15)  =  399     simplify

4T + 5T + 75   =   399

9T + 75   = 399    subtract 75 from both sides

9T   =   324         divide both sides by 9

T  =  36  (dollars)

-7-2(8m+6) = -6m+31     simplify the left side

-7 - 16m - 12   =   -6m + 31

-19 - 16m  =  - 6m  +  31      add 6m, 19 to both sides

-10 m   =  50        divide both sides by -10

m  =  -50 / 10   =   -5

thank you 

4x+3-2x=17+2x-7     simplify

2x + 3   =   10 + 2x 

2x + 3   =  2x + 10

Note that adding different quantities to 2x  can never make the equation equal......thus.....no solutions exist

I think we have

18 / (x^2-3x) -  6/(x-3) = 5/x    

18 / [ x (x - 3) ]  -  6 / (x - 3)   =  5/ x           the LCD  is  x(x-3)

Multiplying through by the LCD produces

18  - 6x  = 5 (  x - 3)

18 - 6x  = 5x - 15      add 6x, 15 to both sides

33 =  11x          divide both sides by 11

3 = x        thus.......there is no solution since this value for x makes an original denominator  = 0

 -30 + 6p = 7(p-5)       simplify the right side

-30 + 6p  = 7p  -  35      subtract 6p from both sides, add 35 to both sides

5  =  p

No absolutely no.

6x/x+4 + 4 = 2x-2/x-1

Do you mean     

6x/(x+4) + 4 = (2x-2)/(x-1)     ?

\(\frac{6x}{(x+4)} + 4 = \frac{(2x-2)}{(x-1)}\\     LCD=(x+4)(x-1)\\ (x+4)(x-1)[\frac{6x}{(x+4)} + 4 ]=(x+4)(x-1)[ \frac{(2x-2)}{(x-1)}]\\     \frac{6x(x+4)(x-1)}{(x+4)} + 4 (x+4)(x-1)= \frac{(2x-2)(x+4)(x-1)}{(x-1)}\\     6x(x-1)+ 4 (x+4)(x-1)=(2x-2)(x+4)\\ 6x^2-6x+ 4 (x^2+3x-4)=2x^2+6x-8\\ 6x^2-6x+ 4x^2+12x-16=2x^2+6x-8\\ 8x^2-8=0\\ x^2-1=0\\ (x-1)(x+1)=0\\ x=\pm1 \)

Edit:

x cannot equal 1 becasue it would make the denominator in the question = 0 

so the answer is 

x = -1

Thanks for pointing this out to me CPhill   

I should have noted at the very beginning that x cannot equal  1 or -4 !!

Mistakes like this need to be carefully guarded against !!

Remember.....dividing two numbers with the same sign produces a positive result. Dividing two numbers with opposite signs produces a negative result.......so.....

-54   =  54   =   6

 -9         9

V_sphere  =   (4/3) * pi * radius^3    ....   so  ....

250  =  (4/3) * pi * radius^3            divide both sides by (4/3) pi

250 / [ (4/3) pi ]   =   radius^3        take the cube root of each side

∛ [ 250 /  (4/3 pi) ]   =  r   ≈  3.9   inches

You don't get to be a moderator just by asking..

((5-1)*3) - 3

nope sorry lol (idk if you can)