Maths questions help please! (1 Viewer)

[jackerino]

Im stuck on a few questions of my maths assignment, quite a few actually. Please help someone!!??!

Q2. Simplify (n-1)C2+nC2 (Its a combination thing). State any restrictions that must be placed on n



Q7. Heres how to draw it: Draw a scalene triangle, and label the bottom right angle A, and clockwise add B and C to other angles. Now on BA there is a midpoint T. Connect T to C. ALso, on BC there is a point P. Connect A to P. Now AP intersects TC at E. AE is perpendicular to CT. TC bisects angle C. Connect E with M.
i) Prove triangle ACE is congruent to triangle PCE ( I cant find a third reason, only two)
ii) Explain why AE= EP (Matching sides congruent triangles from i)
iii) Hence prove EM is parallel to PB
This question is damn easy lolz I just cant seem to get the right result, how stupid of me -_-




Q15. Here's how to draw diagram: Draw a right angled triangle ABC, labelled clockwise from A at the top. Make C face west of the page. Now with AB as a common side, draw a reflection of the triangle, except extend the hypotenuse and AB further so A reaches to D, which is past B, and the hypotenuse reaches to E, such that the formed angle of ADE =90 degrees. Now connect C to E. CE intersects BD at F. Overall, triangle ABC is similar to triangle AD, and AB=a units. The ratio of similarity of AB:AD is k:l

Here are the questions:
a) Prove triangle FBC is similar to triangle FDE
b) find the ratio of the lengths BF to FD
c) SHow that FD=(al(l-k))/(k(l+k))
d) Given AB:AD=2:3, and FD is an integer, find the possible values of A.


Q16.

a) Prove tan(a+b+c)=(tana+tanb+tanc-tan(a)tan(b)tan(c))/(1-tanatanb-tanatanc-tanbtanc)
i)If a, b and c are the angles of a triangle, explain why tana+tanb+tanc-tanatanbtanc=0
b)Given tana/2=tanba/3=tanc/4=k, show that k=sqrt(6)/4
c) Hence find the size of a in degrees and minutes




SORRY ITS SO BIG! Its just super hard and im a bit confused. PLease help with working out!

 

[Demise]

Question 3
15/(2x - 1) >= x

15 >= x(2x - 1)

15 >= 2x^2 - x

0 >= 2x^2 - x - 15

0 >= (2x + 5)(x - 3)

0 >= x - 3
3 >= x

0 >= 2x + 5
-2x >= 5
x <= 5/-2

Hopefully that's right!

 

[jackerino]

Question 3
15/(2x - 1) >= x

15 >= x(2x - 1)

15 >= 2x^2 - x

0 >= 2x^2 - x - 15

0 >= (2x + 5)(x - 3)

0 >= x - 3
3 >= x

0 >= 2x + 5
-2x >= 5
x <= 5/-2

Hopefully that's right!

Youve got to jmultiply by (2x-1)^2, not just 2x-1. The result is a cubic, which I cant seem to solve.

 

[FTW]

Youve got to jmultiply by (2x-1)^2, not just 2x-1. The result is a cubic, which I cant seem to solve.

You don't have to multiply by (2x-1)^2
Just consider when the denominator is equal to zero ie, when x is 1/2
Then x =/= 1/2
Then solve for equals, and test points in the original equation.

 

[jackerino]

I fouund my mistake. its all good.

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x@plus;5)(x-3)\geq 0 \\\\ Ta -da" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x+5)(x-3)\geq 0 \\\\ Ta -da" title="\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x+5)(x-3)\geq 0 \\\\ Ta -da" /></a>

 

[jackerino]

Question 3
15/(2x - 1) >= x

15 >= x(2x - 1)

15 >= 2x^2 - x

0 >= 2x^2 - x - 15

0 >= (2x + 5)(x - 3)

0 >= x - 3
3 >= x

0 >= 2x + 5
-2x >= 5
x <= 5/-2

Hopefully that's right!

Yea u got one result lolz. Then there's (1/2)<=(x)<=(3)

 

[Timske]

1/2 < x , since x=/=1/2

 

[jackerino]

<a href="http://www.codecogs.com/eqnedit.php?latex=\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x@plus;5)(x-3)\geq 0 \\\\ Ta -da" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x+5)(x-3)\geq 0 \\\\ Ta -da" title="\dpi{100} \frac{15}{2x-1}\geq x ~~,~~ x\neq \frac{1}{2} \\\\ 15(2x-1)\geq x(2x-1)^2 \\\\15(2x-1)-x(2x-1)^2\geq 0 \\\\ (2x-1)\left [ 15-x(2x-1) \right ]\geq 0 \\\\-(2x-1)(2x^2-x-15)\geq0 \\\\ -(2x-1)(2x+5)(x-3)\geq 0 \\\\ Ta -da" /></a>

Woah thats nice and quick Timske, I went thru a whole polynomial division thing LOL thanks for the quicker method!

 

[jackerino]

1/2 < x , since x=/=1/2

Oh yea good point.

 

[Timske]

Woah thats nice and quick Timske, I went thru a whole polynomial division thing LOL thanks for the quicker method!

 

[jackerino]

TImske could you by any chance help me do Q16 from my original post? Or if you're busy, a quick outline perhaps please? Like, idk how to apply compound angle formula to three angles LOL

 

[Timske]

TImske could you by any chance help me do Q16 from my original post? Or if you're busy, a quick outline perhaps please? Like, idk how to apply compound angle formula to three angles LOL

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(a@plus;b)=\frac{tana@plus;tanb}{1-tanatanb} \\\\ tan(a@plus;b@plus;c)=tan(a@plus;(b@plus;c))=\frac{tana@plus;tan(b@plus;c)}{1-tanatan(b@plus;c)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(a+b)=\frac{tana+tanb}{1-tanatanb} \\\\ tan(a+b+c)=tan(a+(b+c))=\frac{tana+tan(b+c)}{1-tanatan(b+c)}" title="tan(a+b)=\frac{tana+tanb}{1-tanatanb} \\\\ tan(a+b+c)=tan(a+(b+c))=\frac{tana+tan(b+c)}{1-tanatan(b+c)}" /></a>


That should help

 

[Sanjeet]

TImske could you by any chance help me do Q16 from my original post? Or if you're busy, a quick outline perhaps please? Like, idk how to apply compound angle formula to three angles LOL

for i) if a b c are angles of a triangle, they add up to 180. hence tan(a+b+c) = tan(180) = 0
from the right hand side, the denominator can't be zero, so the numerator must equal zero

 

[jackerino]

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(a@plus;b)=\frac{tana@plus;tanb}{1-tanatanb} \\\\ tan(a@plus;b@plus;c)=tan(a@plus;(b@plus;c))=\frac{tana@plus;tan(b@plus;c)}{1-tanatan(b@plus;c)}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(a+b)=\frac{tana+tanb}{1-tanatanb} \\\\ tan(a+b+c)=tan(a+(b+c))=\frac{tana+tan(b+c)}{1-tanatan(b+c)}" title="tan(a+b)=\frac{tana+tanb}{1-tanatanb} \\\\ tan(a+b+c)=tan(a+(b+c))=\frac{tana+tan(b+c)}{1-tanatan(b+c)}" /></a>


That should help

Oh yay! thanks Next bit confusing though, if (tana)/2=(tanb/3)=(tanc/4)=k, show k=(sqrt6)/4, then hence find the size of angle a in degrees and minutes. Hmm so sorry if im inconveniencing you guys!

 

[Demise]

Yea u got one result lolz. Then there's (1/2)<=(x)<=(3)

Actually I got 2 results there... 3 is greater than or equal to x.
x is less than or equal to -2.5.