All Answers 352881 Answers

i need the answer and solution preferably please.

if  \(\large{ \tan(120^\circ-x)=\dfrac{\sin(120^\circ) -\sin(x)} {\cos(120^\circ) -\cos(x)} }\), where \(\large{0^\circ < x < 180^\circ}\), compute \(x\)

\(\text{Let $ \mathbf{s} = \sin(120^\circ)= \sin(60^\circ)=\dfrac{\sqrt{3}} {2} $} \\ \text{Let $ \mathbf{c} = \cos(120^\circ)=-\cos(60^\circ)=-\dfrac{1} {2} $} \\ \)

LHS:

\(\begin{array}{|rcll|} \hline \tan(120^\circ-x) &=& \dfrac{\sin(120^\circ-x)} {\cos(120^\circ-x)} \\\\ &=& \dfrac{\sin(120^\circ)\cos(x)-\cos(120^\circ)\sin(x)} {\cos(120^\circ)\cos(x)+\sin(120^\circ)\sin(x)} \\\\ &=& \dfrac{s\cdot \cos(x)-c\cdot\sin(x)} {c\cdot\cos(x)+s\cdot\sin(x)} \\ \hline \end{array}\)

RHS:

\(\begin{array}{|rcll|} \hline \dfrac{\sin(120^\circ) -\sin(x)} {\cos(120^\circ) -\cos(x)} \\\\ &=& \dfrac{ 2\cos\left(\dfrac{120^\circ+x}{2}\right) \sin\left(\dfrac{120^\circ-x}{2}\right) } {-2\sin\left(\dfrac{120^\circ+x}{2}\right) \sin\left(\dfrac{120^\circ-x}{2}\right)} \\\\ &=& -\dfrac{ \cos\left(\dfrac{120^\circ+x}{2}\right) } { \sin\left(\dfrac{120^\circ+x}{2}\right) } \\\\ &=& -\dfrac{ \cos\left(60^\circ+\dfrac{x}{2}\right) } { \sin\left(60^\circ+\dfrac{x}{2}\right) } \\\\ &=& -\dfrac{\cos(60^\circ)\cos\left(\dfrac{x}{2}\right) -\sin(60^\circ)\sin\left(\dfrac{x}{2}\right) } {\sin(60^\circ)\cos\left(\dfrac{x}{2}\right) +\cos(60^\circ)\sin\left(\dfrac{x}{2}\right) } \\\\ &=& \dfrac{\sin(60^\circ)\sin\left(\dfrac{x}{2}\right) -\cos(60^\circ)\cos\left(\dfrac{x}{2}\right) } {\sin(60^\circ)\cos\left(\dfrac{x}{2}\right) +\cos(60^\circ)\sin\left(\dfrac{x}{2}\right) } \\\\ &=& \dfrac{s\cdot \sin\left(\dfrac{x}{2}\right) +c\cdot \cos\left(\dfrac{x}{2}\right) } {s\cdot \cos\left(\dfrac{x}{2}\right) -c\cdot \sin\left(\dfrac{x}{2}\right) } \\ \hline \end{array}\)

\(\begin{array}{|lcll|} \hline \dfrac{s\cdot \cos(x)-c\cdot\sin(x)} {c\cdot\cos(x)+s\cdot\sin(x)} = \dfrac{s\cdot \sin\left(\dfrac{x}{2}\right) + c\cdot \cos\left(\dfrac{x}{2}\right) } {s\cdot \cos\left(\dfrac{x}{2}\right) - c\cdot \sin\left(\dfrac{x}{2}\right) } \\\\ (s\cdot \cos(x)-c\cdot\sin(x)) \left(s\cdot \cos\left(\dfrac{x}{2}\right) - c\cdot \sin\left(\dfrac{x}{2}\right) \right) \\ = (c\cdot\cos(x)+s\cdot\sin(x)) \left( s\cdot \sin\left(\dfrac{x}{2}\right) + c\cdot \cos\left(\dfrac{x}{2}\right) \right) \\\\ s^2\cos(x)\cos\left(\dfrac{x}{2}\right) - sc\cos(x)\sin\left(\dfrac{x}{2}\right)- sc\sin(x)\cos\left(\dfrac{x}{2}\right)+c^2\sin(x)\sin\left(\dfrac{x}{2}\right) \\ =c^2\cos(x)\cos\left(\dfrac{x}{2}\right) + sc\cos(x)\sin\left(\dfrac{x}{2}\right)+ sc\sin(x)\cos\left(\dfrac{x}{2}\right)+s^2\sin(x)\sin\left(\dfrac{x}{2}\right) \\\\ \cos(x)\cos\left(\dfrac{x}{2}\right)(s^2-c^2)-\sin(x)\sin\left(\dfrac{x}{2}\right)(s^2-c^2)-2sc\cdot \cos(x)\sin\left(\dfrac{x}{2}\right)-2sc\cdot \sin(x)\cos\left(\dfrac{x}{2}\right) = 0 \\\\ (s^2-c^2)\left(\underbrace{ \cos(x)\cos\left(\dfrac{x}{2}\right)-\sin(x)\sin\left(\dfrac{x}{2}\right) }_{=\cos\left(x+\dfrac{x}{2}\right)} \right) -2sc\cdot\left(\underbrace{ \cos(x)\sin\left(\dfrac{x}{2}\right)+\sin(x)\cos\left(\dfrac{x}{2}\right) }_{=\sin\left(x+\dfrac{x}{2}\right)} \right) = 0 \\\\ (s^2-c^2)\cos\left(x+\dfrac{x}{2}\right) - 2sc\cdot \sin\left(x+\dfrac{x}{2}\right) = 0 \\\\ 2sc\cdot \sin\left(x+\dfrac{x}{2}\right) = (s^2-c^2)\cos\left(x+\dfrac{x}{2}\right) \\\\ 2sc\cdot \sin\left(\dfrac{3x}{2}\right) = (s^2-c^2)\cos\left(\dfrac{3x}{2}\right) \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{s^2-c^2} {2sc} \quad | \quad s^2-c^2 = \dfrac34-\dfrac14 = \dfrac12 \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{\dfrac12} {2sc} \quad | \quad 2sc = 2\dfrac{\sqrt{3}}{2}\left(-\dfrac12\right)=-\dfrac{\sqrt{3}}{2} \\\\ \tan\left(\dfrac{3x}{2}\right) = \dfrac{\dfrac12} {-\dfrac{\sqrt{3}}{2}} \\\\ \tan\left(\dfrac{3x}{2}\right) = -\dfrac{\sqrt{3}}{3} \\\\ \dfrac{3x}{2} = \arctan(-\dfrac{\sqrt{3}}{3}) +n\cdot 180^\circ \qquad n\in \mathbb{Z} \\\\ x= \dfrac{2}{3}\cdot \arctan\left(-\dfrac{\sqrt{3}}{3}\right) +\dfrac{2}{3}\cdot n\cdot 180^\circ \\\\ x= \dfrac{2}{3}\cdot (-30^\circ) + n\cdot 120^\circ \\\\ x= -20^\circ + n\cdot 120^\circ \qquad n = 1 \quad (0^\circ < x < 180^\circ) \\\\ x= -20^\circ + 120^\circ \\\\ \mathbf{x= 100^\circ } \\ \hline \end{array}\)

Yes....I understand PEDMAS   you must follow pedmas to solve these equations.....

Sorry...I am not sure what you mean by 'inverse operations'....... can you teach me what that means?

2(3x-7) + 18 = 10

6x-14+18 = 10

6x + (-14+18) = 10

6x + 4 = 10

6x + (4-4) = 10 -4

6x = 6

x = 1  

1) Interest rate conversion:

Effective annual rate =[1 + 1.5%]^12

                                  =[1 + 0.015]^12

                                  =[1.015]^12

                                  =1.19561817...  - 1 x 100

                                  =~19.6% - effective annual rate.

2)   FV =PV x [1 + R]^N

            =PV x [1 + 1.5%]^(15*12)

            =$100,000 x [1.015]^180

            =$100,000 x  14.58436768913............

            =$1,458,437 - Growth in this investment after 15 year @ 1.5% comp.monthly

The answer is "a".

Find all real values of a for which the equation \(\mathbf{(x^2 + a)^2 + a = x}\) has four real roots.

My attempt:

I assume there are 4 distinct real roots. So there are 3 local maxima/minima.

We find the local maxima/minima by differentiation. Maxima/minima occur when f'(x) = 0

1. Differentiation:

\(\begin{array}{|rcll|} \hline y &=& (x^2 + a)^2 -x+a \\ y' &=& 2(x^2+a)\cdot 2x-1 \\ 0 &=& 4x(x^2+a) - 1 \\ 4x(x^2+a) &=& 1 \quad | \quad : 4 \\ x(x^2+a) &=& \frac14 \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

The zero set of discriminant of the cubic  \(\mathbf{Ax^{3}+Bx^{2}+Cx+D\,}\) has discriminant
  \(\mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD\,.}\)
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

\(\begin{array}{|rcll|} \hline \mathbf{Ax^{3}+Bx^{2}+Cx+D\,} \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \quad | \quad A=1,\ B=0,\ C=a,\ D=-\frac14 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD } &>& 0 \quad | \quad \text{there are three distinct real numbers} \\ -4a^3 -27\left(-\frac14\right)^2 &>& 0 \\ -4a^3 -\frac{27}{16} &>& 0 \\ -4a^3 &>& \frac{27}{16} \quad | \quad : -4 \\ a^3 &<& -\frac{27}{64} \\ a^3 &<& -\frac{3^3}{4^3} \\ \mathbf{ a } &\mathbf{<}& \mathbf{-\frac{3}{4}} \\ \hline \end{array}\)

\(\mathbf{(x^2 + a)^2 + a = x}\) has four distinct real roots if  \(\mathbf{ a<-\frac{3}{4}}\)

yes, I saw that, thank you so much for your time...I do apreciate!..as always!.. 

top right

Hi Melody,

According to the handbook, "b" can never be negative. It can only be either a fraction between 0 and 1, or greater than 1. Okay, so this I accept. There is an example in the book on a similar sum, but it is not clear to me. By the way it looks, it appears the "a" is disregarded and "b" is calculated. I followed that aproach and found "b" to be \(1 \over2\),

I then substituted this also into the equation and found "a" to be -1. But I'm not sure if that was the right way to go about it?

ok but you should check your own answers by plugging in the value that you get into both sides of the equation and seeing if the 2 sides are the same.

You do not need other people to check questions like this, you can, and should, do it for yourself. 

If you want other people to check other types of questions for you  (and you already have your answer) you should state that in your question. 

\(1=a+q\\ q=1-a\)

\(\frac{-1}{3}=ab^{-1}+q\\ \frac{-1}{3}=\frac{a}{b}+q \qquad \qquad b\ne0 \\ \frac{-1}{3}=\frac{a}{b}+1-a\\ 3b(\frac{-1}{3})=3b(\frac{a}{b}+1-a)\\ -b=3a+3b-3ab\\ 0=3a+4b-3ab \qquad \text{a and b both equal 0 or}\\ -3a(1-b)=4b\\ 3a(b-1)=4b\\ a=\frac{4b}{3(b-1)} \qquad \qquad b\ne1\\ \)

If    b =1    then

\(g(x)=a+q\\ g(x)=a+1-a\\ g(x)=1 \)

but then point G is not on the curve so  b cannot equal 1

So we have

.\(q=1-a\\ a=\frac{4b}{3(b-1)}\\ where \;\; b\ne1 \;\;\;\; b\ne0\)

PLUS, I am not so sure that b can be negative........   

The graph would be very broken if b was negative, I am not sure how much that matters.

Maybe another mathematician would like to comment here. 

Anyway ,...There is no single solution for this.

Here is the graph.

If a, b, c and d are positive real numbers such that
log_a  b= 8/9,
log_b  c= -(3/4),
log_c d = 2,
find the value of log_d (abc)

\(\begin{array}{|rclcl|} \hline \log_a(b) &=& \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{\log_b(b)}{\log_b(a)} \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{\log_b(b)}{\log_b(a)} \quad | \quad \log_b(b) = 1 \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \dfrac{1}{\log_b(a)} \quad | \quad \log_a(b)\log_b(a)=1 \\\\ & & \dfrac{\log_d(b)}{\log_d(a)} &=& \log_a(b) \\\\ & & \dfrac{\log_d(a)} {\log_d(b)}&=& \dfrac{1}{\log_a(b)} \\\\ & & \mathbf{\log_d(a)} & \mathbf{=} & \mathbf{ \dfrac{\log_d(b)}{\log_a(b)} } \qquad (1) \\ \hline \log_b(c) &=& \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{\log_c(c)}{\log_c(b)} \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{\log_c(c)}{\log_c(b)} \quad | \quad \log_c(c) = 1 \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \dfrac{1}{\log_c(b)} \quad | \quad \log_b(c)\log_c(b)=1 \\\\ & & \dfrac{\log_d(c)}{\log_d(b)} &=& \log_b(c) \\\\ & & \dfrac{\log_d(b)}{\log_d(c)} &=& \dfrac{1}{\log_b(c)} \\\\ & & \mathbf{\log_d(b)} &\mathbf{=}& \mathbf{\dfrac{\log_d(c)}{\log_b(c)}} \qquad (2) \\ \hline && \log_c(d) &=& \dfrac{\log_d(d)}{\log_d(c)} \quad | \quad \log_d(d) = 1 \\\\ && \log_c(d) &=& \dfrac{1}{\log_d(c)} \\\\ & & \mathbf{\log_d(c)} &\mathbf{=}& \mathbf{\dfrac{1}{\log_c(d)}} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \log_d(abc) &=& \log_d(a)+\log_d(b)+\log_d(c) \quad | \quad \mathbf{\log_d(a)=\dfrac{\log_d(b)}{\log_a(b)} } \qquad (1) \\ &=& \dfrac{\log_d(b)}{\log_a(b)}+\log_d(b)+\log_d(c) \\ &=& \log_d(b) \left( \dfrac{1}{\log_a(b)}+1 \right) +\log_d(c) \quad | \quad \mathbf{\log_d(b)=\dfrac{\log_d(c)}{\log_b(c)}} \qquad (2) \\ &=& \dfrac{\log_d(c)}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) +\log_d(c) \\ &=&\log_d(c)\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right) \quad | \quad \mathbf{\log_d(c)=\dfrac{1}{\log_c(d)}} \qquad (3) \\ \mathbf{\log_d(abc)} & \mathbf{=} & \mathbf{\dfrac{1}{\log_c(d)}\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\log_d(abc)} & \mathbf{=} & \mathbf{\dfrac{1}{\log_c(d)}\left( \dfrac{1}{\log_b(c)} \left( \dfrac{1}{\log_a(b)}+1 \right) + 1 \right)} \\\\ & = & \dfrac{1}{2}\left( \dfrac{1}{ -\dfrac{3}{4} } \left( \dfrac{1}{\dfrac{8}{9}}+1 \right) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-4}{3} \Big( \dfrac{1}{8}+1 \Big) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-4}{3} \left( \dfrac{17}{8} \right) + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-17}{6} + 1 \right) \\\\ & = & \dfrac{1}{2}\left( \dfrac{-11}{6} \right) \\\\ & \mathbf{=} & -\mathbf{\dfrac{11}{12}} \\ \hline \end{array}\)

Thanks!!

\(\cos(x) - \cos(2x) = 0\\ \cos(x) = \cos(2x)\\ \cos(x) = 2\cos^2(x) -1\\ 2\cos^2(x) - \cos(x) - 1 = 0\)

\(u = \cos(x)\\ 2u^2 - u - 1 = 0\\ (2u+1 )(u -1) = 0\\ u = 1,~-\dfrac 1 2\)

\(\cos(x) = 1 \Rightarrow x = 2k\pi, ~k \in \mathbb{Z}\\ \cos(x) = -\dfrac 1 2 \Rightarrow x = \dfrac{2\pi}{3}+2k\pi,~\dfrac{4\pi}{3}+2k\pi,~k\in \mathbb{Z}\)

\(x = \left \{0,~\dfrac{2\pi}{3},~\dfrac{4\pi}{3}\right\} + 2\pi k,~k \in \mathbb{Z}\)

\(\log_x(y^2) + \log_y(x^2) = \\ 2 \log_x(y) + 2\log_y(x) = \\ 2( \log_x(y) + \log_y(x) ) = \\ 2 \cdot 3 = 6\)

I am not sure which 1 you are referring to, but....

\(\frac{sinxcosx-sinx}{cos^2x-1}\\~\\ \qquad \text{Factorise the top}\\~\\= \frac{-sinx(-cosx+1)}{cos^2x-1}\\~\\ \qquad Sin^2x+cos^2x=1 \qquad so\\~\\ =\frac{-sinx(1-cosx)}{-sin^2x}\\~\\ \qquad \text{cancel}\\~\\ =\frac{1-cosx}{sinx}\\~\\\)

\(x^2 - 2x - 15 = (x+3)(x-5)\\ \text{roots are }x = -3,~5\)

\(-(x + 7)-5 = 4(x+3) - 6x\\ -x - 7 - 5 = 4x + 12 - 6x \\ -x - 12 = -2x + 12\\ -x + 2x - 12 = 12\\ -x + 2x = 24\\ x = 24\)

1) X^2  - 7X + 12

Answer i wrote: (X - 4) (X - 3)

correct

2) X^2 + 5X - 6

Answer i wrote: (X + 6) (X - 1)

correct

3) 3X^2 - X - 2

Answer i wrote: (3x - 2) (x + 1)

incorrect (check signs)

4) 25X^2 - 30X + 9

Answer i wrote: (5X - 3) (5X -3)

correct

5) 36X^2 - 9Y^2

Answer i wrote: Square root both 36x^2 and 9y^2

=(6X - 3Y) (6X + 3Y) 

correct 

3)

3X^2 - X - 2

Answer i wrote: (3x - 2) (x + 1) no!

3X² - X - 2 = (3x + 2) (x - 1)