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Mathematics: Post your doubts here!

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Re: Maths help available here!!! Stuck somewhere?? Ask here!

The answer must be 108 and not 10.8 which is too less.
 

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Re: Maths help available here!!! Stuck somewhere?? Ask here!

in the triangle abc, ab=12 cm, angle bac=60 degree and angle acb=45degree. Find the exact length of bc PLEASE DOO IT FAST
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

Sine rule:
bc/sin60=12/sin45
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

By sine rule, BC/sin bac = AB/sin acb
BC / sin 60 = 12 / sin 45
BC / (√ 3/2) = 12 / (1/√ 2)
BC * (2/√ 3) = 12 * √ 2
BC = (12 * √ 2 * √ 3) / 2
BC = 6√ 6
Hpe it hlps...
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

please answer this question, with angles.
2Sin(theta) + 8Cos^2(theta) = 5
 
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babamama99 said:
please answer this question, with angles.
2Sin(theta) + 8Cos^2(theta) = 5
 

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Re: Maths help available here!!! Stuck somewhere?? Ask here!

ma prblm's not exactly being stuck sumwhere... itx a whole chapter..! cud any one here please jxt explain to me the basics of the topic circular measure??? in "core maths"(buk) by Bostok n Chandler itx chapter 10. i noe itx really stupid to be askng this but i srsly dunno get this topic.
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

Mubarka said:
ma prblm's not exactly being stuck sumwhere... itx a whole chapter..! cud any one here please jxt explain to me the basics of the topic circular measure??? in "core maths"(buk) by Bostok n Chandler itx chapter 10. i noe itx really stupid to be askng this but i srsly dunno get this topic.
Some videos that i loved for this chapter :)
KhanAcademy
AlRichards314
ExamSolutions
MathsTutorBiz
MathStudentTeacher

The ones from Khan Academy, Exam Solutions and Math Student Teacher were awesome !! Hope they'd help you !! :)
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

got a very weird question here:

hx^2+hx+2h=3

the equation has real roots, find h

the answer at the back of the book is: 0<h<4/3

i just don't get it, that must be the wrong answer, I keep getting something else

somebody solve this :(

P.S the '<' stands for 'is lesser then and equal to, and vice versa
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!
 

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Re: Maths help available here!!! Stuck somewhere?? Ask here!

so brother anser at bak of book must be wrong it shud be 0<h<12/7
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

Answer at the back of the book is wrong.
 

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Re: Maths help available here!!! Stuck somewhere?? Ask here!

@Hassam: Sorry, didn't see your solution because it was on a new page. :S
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

I have a doubt here, hope someone can help me

i couldnt get the answer as shown in marking scheme

Thanks!
 

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buddy my advice for solving modular inequalities.....if u have modulus on both sides u can take square and solve.....secondly if u have a situation where u have modulus on one side and a number on the other.....u cn only square if that number is positive....else u ll get extraneous roots.......thirdly ....the case in this question....u hav moduls on one side and a non-modular expression on other side involving X.....in this case since u cnt be sure that 2-3x is Positive always u shud not take square....cos u ll get an extraneous root.....which u ll have to cancel by observation..and u cn forget,,,,, so in these situation u must sketch .....stick with these 4 basic rules.....m sure u ll solve any question on this topic.....well so go for OR3 option in this case.....i ll be psoting a pictorial solution soon
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!


so u see a critical point at 2-3x=-(x-3)
 

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Re: Maths help available here!!! Stuck somewhere?? Ask here!

hassam said:
so u see a critical point at 2-3x=-(x-3)

thanks! my tutor told me sketching graph is the last option, and seems like couldnt get it using normal way, so i should just sketch it, Thanks! :good:
 
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Re: Maths help available here!!! Stuck somewhere?? Ask here!

Assalamoalaikum!
I have one question: What is the point of Logarithms? I mean, what benefit do we get from them? We can simply use the indices instead of logs........
 
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