All Answers

+118723

Inverse tangent is atan

MelodyFeb 10, 2015

+118723

Saseflower, you have subtracted b on the left sides and divided by b on the right side.

You can't do that :))

----------------------------------------------------

a+b+y+s+e=sb , ae-ys=yy , y+b=y

Well if y+b=y then b must equal 0

a+0+y+s+e=s*0 , ae-ys=yy

a+y+s+e=0 , ae-ys=yy

Mult both sides by y

ay+yy+sy+ey=0 ae=yy+ys

ay+ae+ey=0

Now I don't know

MelodyFeb 10, 2015

+118723

Unless you are studying at a very advanced level I do not think that this question is correct.

MelodyFeb 10, 2015

+26400

(2) if b=square root 13 ,and a+c=4 what is the area of the triangle ABC

$$\boxed{\small{\text{(1) The shape of the triangle by Heron's formula:$\quad A^2=s(s-a)(s-b)(s-c) \quad s = \frac{a+b+c}{2}$
}} }\\\\
\small{\text{$s =\frac{a+c+b}{2} \quad | \quad a+c=4
$
}}\\
\small{\text{
$s = \frac{4+b}{2} = 2 + \frac{b}{2}\qquad
\boxed{s=2 + \frac{b}{2}
} $
}}\\\\
\small{\text{
$
A^2=s(s-a)(s-b)(s-c) = \textcolor[rgb]{1,0,0}{s(s-b)} (s-a)(s-c) $
}}
}}\\
\small{\text{
$
\textcolor[rgb]{1,0,0}{s(s-b)}=( 2+\frac{b}{2} )(2+\frac{b}{2}-b) =(2+\frac{b}{2})(2-\frac{b}{2})=4-\frac{b^2}{4} \quad | \quad b^2=13
$
}}\\
\small{\text{
$
\textcolor[rgb]{1,0,0}{s(s-b)} = 4-\frac{13}{4} =\frac{16-13}{4} = \frac{3}{4}$
}}\\\\
\small{\text{
$
A^2=\textcolor[rgb]{1,0,0}{s(s-b)} (s-a)(s-c) = \textcolor[rgb]{1,0,0}{ \frac{3}{4} } (s-a)(s-c) $
}}\\
\small{\text{
$
A^2=\frac{3}{4}(s-a)(s-c)= \frac{3}{4}(s^2-sc-sa+ac)
=\frac{3}{4}(s^2-s(a+c)+ac) \quad | \quad a+c=4
$
}}\\
\small{\text{
$
A^2=\frac{3}{4}( \textcolor[rgb]{1,0,0}{s^2-4s} +ac)$
}}\\
\small{\text{
$
\textcolor[rgb]{1,0,0}{s^2-4s} = (2+ \frac{b}{2} )^2-4(2+\frac{b}{2}
)=4+2b+ \frac{b^2}{4} -8-2b=-4+\frac{b^2}{4} =\frac{-16+13}{4}= -\frac{3}{4}
$
}}\\\\
\small{\text{
$
A^2=\frac{3}{4}( \textcolor[rgb]{1,0,0}{ -\frac{3}{4}
} +ac) $
}}\\
\small{\text{
$
A^2*\frac{4}{3} -ac + \frac{3}{4} = 0$
}}\\$$

$$\boxed{\small{\text{(2) The shape of the triangle :$\quad
A= \dfrac {ac \sin{(B)} } {2} \qquad ac = \dfrac{2A}{\sin{(B)}}
$
}} }\\\\
\small{\text{
$
A^2*\frac{4}{3} -ac + \frac{3}{4} = 0$
}} }\\
\small{\text{
$
A^2*\frac{4}{3} -\frac{2A}{\sin{(B)}} + \frac{3}{4} = 0$
}} }\\\\
\small{\text{
$
A_{1,2}=
\dfrac
{
\frac{2}{\sin{(B)}}
\pm\sqrt{
\left( \frac{2}{ \sin{(B)} } \right)^2-4*\frac{4}{3}*(\frac{3}{4} ) }
}
{ 2*\frac{4}{3} }
=
\dfrac
{
\frac{2}{\sin{(B)}}
\pm 2\sqrt{
\frac{1}{ \sin^2{(B)} }-1 }
}
{ \frac{8}{3} }
=
\dfrac
{
\frac{2}{\sin{(B)}}
\pm 2 \frac{ \cos{(B)} } { \sin{(B)} } }
{ \frac{8}{3} }
$
}}\\
\small{\text{
$
A_{1,2}
=
\frac{3}{4}
\left(
\frac{1}{\sin{(B)}}
\pm \frac{ \cos{(B)} } { \sin{(B)} }
\right)
=
\frac{3}{4}
\left(
\frac{1\pm\cos{(B)} }{\sin{(B)}}
\right)
$
}}\\\\
\small{\text{
$
A_{1}
=
\frac{3}{4}
\left(
\frac{1+\cos{(B)} }{\sin{(B)}}
\right)
\quad | \quad \cos{(B)} = - \frac{1}{2} \quad \sin{(B)} =\sqrt{ 1-\cos^2{ (B) }} = \frac{\sqrt{3}}{2}
$
}}\\
\small{\text{
$
A_{1}=\frac{3}{4}*\frac{1+ (-\frac{1}{2}) }{ \frac{\sqrt{3}}{2} }
=\frac{3}{4}*\frac{2}{ \sqrt{3} } *\frac{1}{2}
=\frac{3}{4}*\frac{2*\sqrt{3}}{3} *\frac{1}{2}
= \frac{1}{4}*\sqrt{3}
$
}}\\
\small{\text{
$
\boxed{A_{1}= \frac{1}{4}\sqrt{3} $ no solution!$}
$
}}\\\\
\small{\text{
$
A_{2}
=
\frac{3}{4}
\left(
\frac{1-\cos{(B)} }{\sin{(B)}}
\right)
\quad | \quad \cos{(B)} = - \frac{1}{2} \quad \sin{(B)} =\sqrt{ 1-\cos^2{ (B) }} = \frac{\sqrt{3}}{2}
$
}}\\
\small{\text{
$
A_{2}=\frac{3}{4}*\frac{1- (-\frac{1}{2}) }{ \frac{\sqrt{3}}{2} }
=\frac{3}{4}*\frac{2}{ \sqrt{3} }*\frac{3}{2}
=\frac{3}{4}* \frac{2*\sqrt{3}}{3} *\frac{3}{2}
= \frac{3}{4}*\sqrt{3}
$
}}\\
\small{\text{
$
\boxed{A_{2}= \frac{3}{4}\sqrt{3}}
$
}}$$

heurekaFeb 10, 2015

+118723

i am going to ignore the 29th feb

so there are 365 days in a year.

what is the prob that their birthdays are all different

$$\\P(all \;diff)=1*\frac{364}{365}*\frac{363}{365}*...... \frac{365-n+1}{365}*\frac{365-60+1}{365}\\\\
P(all \;diff)=1*\frac{364}{365}*\frac{363}{365}*...... \frac{366-n}{365}*\frac{306}{365}\\\\
P(all \;diff)=\frac{364!}{305!(365)^{59}}\\\\
P( at least 2 the same)=1-\frac{364!}{305!(365)^{59}}$$

The calc is refusing to calculate this. I'll come back to it if time.

Anyway it is very likely.

MelodyFeb 10, 2015

+118723

Nice work minee and welcome to the web2.0calc forum

MelodyFeb 10, 2015

+118723

Nice presentation Mathematician :)

MelodyFeb 10, 2015

+118723

inverse of -1/4 is -4

MelodyFeb 10, 2015

+118723

(10/3)*2=20/3=6 and 2/3

MelodyFeb 10, 2015

+118723

I think that it is 3,x,9

$$x/3=9/x

x^2=27

x=
\sqrt{27}=3\sqrt3$$

MelodyFeb 10, 2015

+26400

+10

In a triangle ABC, a, b , c are the opposite side of angle A,angle B ,and angle C, and

$${\frac{{\mathtt{cosB}}}{{\mathtt{cosc}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{b}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}\right)}}$$

(I) find the angle B

$$\boxed{ \dfrac{ \cos{(B)} }{\cos{(C)}}=-\dfrac{b}{(2a+c)} }
\small{\text{
$\quad $For any triangle:
$
a=b\cos{(C)}+c\cos{(B)}.
$
}}\\\\
\small{\text{
$
(2a+c)\cos{(B)} = -b\cos{(C)} \quad | \quad -b\cos{(C)}= c\cos{(B)} - a
$
}}\\
\small{\text{
$
(2a+c)\cos{(B)} = c\cos{(B)} - a
$
}}\\
\small{\text{
$
(2a)\cos{(B)} = - a
$
}}\\
\small{\text{
$
2\cos{(B)} = - 1
$
}}\\
\small{\text{
$
\textcolor[rgb]{1,0,0}{\cos{(B)} = - \dfrac{1}{2} \qquad B = 120\ensurement{^{\circ}}
}$
}}$$