hectictar

Find all real numbers \(t\) such that \(\frac23 t - 1 \ <\ t + 7\ \le\ -2t + 15\) . Give your answer as an interval.

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\(\frac23 t - 1 \ <\ t + 7\\~\\ \frac23 t - t \ <\ 1 + 7\\~\\ -\frac13t\ <\ 8\\~\\ t\ >\ -24\)

And

\(t + 7\ \le\ -2t + 15\\~\\ t+2t\ \leq\ -7+15\\~\\ 3t\ \leq\ 8\\~\\ t\leq\frac83\)

t  is bigger than  -24  but less than or equal to  8/3

t  must be in the interval  \(\Big(-24,\frac83\Big]\)_

By the Pythagorean Identity,

\(\sin^2(12^\circ)+\cos^2(12^\circ)\ =\ 1\)

                                                      We are given that   cos(12°) = h   so we can substitute  h  in for  cos(12°)

\(\sin^2(12^\circ)+h^2\ =\ 1\)

                                                      Subtract  h²  from both sides of the equation.

\(\sin^2(12^\circ)\ =\ 1-h^2\)

                                                      Because  12°  is in Quadrant I,  sin(12°)  is positive. So take positive sqrt of both sides.

\(\sin(12^\circ)\ =\ \sqrt{1-h^2}\)

By definition of cotangent,

\(\cot(12^\circ)\ =\ \frac{\cos(12^\circ)}{\sin(12^\circ)}\)

                                          Substitute  h  in for  cos(12°)  and substitute  √[ 1 - h² ]  in for  sin(12°)

\(\cot(12^\circ)\ =\ \frac{h}{\sqrt{1-h^2}}\)_

Let the distance between the center of the base and the northwest corner of the base be  y m

Distance between the center of the base and the west side of the base  =  3/2 m  =  1.5 m

Distance between the center of the base and the north side of the base  =  1.5/2 m  =  0.75 m

By the Pythagorean Theorem,

\((1.5)^2+(0.75)^2\ =\ y^2\\~\\ (\frac32)^2+(\frac34)^2\ =\ y^2\\~\\ \frac94+\frac{9}{16}\ =\ y^2\\~\\ \frac{45}{16}\ =\ y^2\\~\\ y^2\ =\ \frac{45}{16}\)

Again by the Pythagorean Theorem,

\(y^2+2^2\ =\ x^2\\~\\ \frac{45}{16}+4\ =\ x^2\\~\\ \frac{109}{16}\ =\ x^2\\~\\ x\ =\ \sqrt{\frac{109}{16}}\\~\\ x\ =\ \frac{\sqrt{109}}{4}\)___

Oh actually this is wrong...I didn't read the question right...it asks for area of triangle ABC  not area of triangle ABD!!

By the Pythagorean Theorem,

(DC)² + (DA)²  =  (AC)²

5² + (DA)²  =  13²

(DA)²  =  13² - 5²

(DA)²  =  144

DA  =  12

By the Pythagorean Theorem,

(DA)² + (BD)²  =  (AB)²

12² + (BD)²  =  15²

(BD)²  =  15² - 12²

(BD)²  =  81

BD  =  9

area of triangle ABD  =  (1/2)(base)(height)

area of triangle ABD =  (1/2)( DA )( BD )

area of triangle ABD =  (1/2)(12)(9)

area of triangle ABD =  54

area of triangle ACD  =  (1/2)(base)(height)

area of triangle ACD  =  (1/2)( DA )( DC )

area of triangle ACD  =  (1/2)( 12 )( 5 )

area of triangle ACD  =  30

area of triangle ABC  =  area of triangle ABD - area of triangle ACD

area of triangle ABC  =  54 - 30

area of triangle ABC  =  24

*edited to fix error*

There is only one solution for  x  when the discrimant is  0 , that is, when...

\((b+\frac1b)^2-4(1)(c)\ =\ 0\\~\\ (b+\frac1b)^2-4c\ =\ 0\\~\\ (b+\frac1b)(b+\frac1b)-4c\ =\ 0\\~\\ b^2+2+\frac{1}{b^2}-4c\ =\ 0\\~\\ b^2+(2-4c)+\frac{1}{b^2}\ =\ 0\\~\\ b^4+(2-4c)b^2+1\ =\ 0\qquad\text{Let}\qquad u=b^2\\~\\ u^2+(2-4c)u+1\ =\ 0\)

There is only one solution for  u, and thus only one positive value of  b ,  when...

\((2-4c)^2-4\ =\ 0\\~\\ (2-4c)^2\ =\ 4\\~\\ 2-4c\ =\ 2\qquad\text{or}\qquad2-4c\ =\ -2\\~\\ c\ =\ 0\phantom{2-4}\qquad\text{or}\qquad c\ =\ 1\)

The non-zero value of  c  is  1 .

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To check this answer, let's find the values of  b  for which   \(x^2+(b+\frac1b)x+1=0\)   has only one solution.

\(x^2+(b+\frac1b)x+1=0\)     has only one solution when....

\((b+\frac1b)^2-4\ =\ 0\\~\\ b^2-2+\frac{1}{b^2}\ =\ 0\\~\\ b^4-2b^2+1\ =\ 0\\~\\ (b^2-1)^2\ =\ 0\\~\\ b\ =\ \pm1\)

There is only one positive value of  b  for which there is one solution to the equation  \(x^2+(b+\frac1b)x+1=0\)

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I got the same answer as you......you can stop crying! You know how to do these questions too.

\(\sqrt{88}\)   can be simplified to   \(2\sqrt{22}\)   because   \(\sqrt{88}\ =\ \sqrt{4\cdot22}\ =\ \sqrt4\cdot\sqrt{22}\ =\ 2\sqrt{22}\)

But it is true that    x = \(\frac{4+\sqrt{88}}{4}\)    

2x²  =  4x + 9

                               Subtract  4x  from both sides of the equation.

2x² - 4x  =  9

                               Subtract  9  from both sides of the equation.

2x² - 4x - 9  =  0

                               By the quadratic formula,

x  =  \({-(-4) \pm \sqrt{(-4)^2-4(2)(-9)} \over 2(2)}\ =\ \frac{4\pm\sqrt{16+72}}{4}\ =\ \frac{4\pm\sqrt{88}}{4}\ =\ \frac{4\pm2\sqrt{22}}{4}\ =\ \frac{2\pm\sqrt{22}}{2}\)

We're given that  x  is positive so....

x  =  \(\frac{2+\sqrt{22}}{2}\)

Now it is in simplified form as   \(\frac{a+\sqrt{b}}{c}\)   where  a,  b,  and  c  are positive integers.

a + b + c   =   2 + 22 + 2   =   26