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10701.

Find the value of Cos 70/sin20+cosec20/sec70-2tan30tan60

Answer» 0
10702.

4sinA+4cosA

Answer»
10703.

Is there any easy method for solving completing square method?

Answer»
10704.

Dgh

Answer»
10705.

1/sinA+1/1-sinA = 2sec²A

Answer»
10706.

What is the value of :√3/sin 10°-1/cos 10°.

Answer»
10707.

If a+b+c=1 a^2+b^2+c^2=2 a^3+b^3+c^3=3 Then find the value of a^4+b^4+c^4.

Answer»
10708.

Exercise 5.3 question 16

Answer»
10709.

Solve the following quadratic equations by factorization.: (1) ( X - 4) (X+2)= 0

Answer» X-4=0. X+2=0X=4. X=-2
Let, (x-4)(x+2)(1)=0X²+2x-4x-8=0X²-2x-8=0X²-(4-2)x-8=0X²-4x+2x-8=0X(x-4)+2(x-4)=0(X+2)(x-4)=0X+2=0. X-4=0X=-2. X=4
10710.

How to use trigonometry identities? I really confuse?

Answer» When the quetion has specific 0,30,45,60,90.
10711.

What is the value of cos 30 degree

Answer» root 3 by 2
10712.

What is the value of sin 30

Answer» 1/2 root
1/2
1/2
10713.

Thals theoram

Answer» The ratio of any two corresponding sides in two equangular triangles is always same
10714.

What type of question paper pattern will come? NCERT or CBSE made questions

Answer» Helen character
10715.

Reasons to contrust Ex11.1 question 2

Answer»
10716.

Easy form to find x valu

Answer»
10717.

Which easy methods to find x value

Answer»
10718.

(sin⁴ theta-cos4 theta) cosec2 theta=2

Answer»
10719.

Show that any positive odd integer is of the form 6q+1,or 6q+3, or 6q+5, where q is some integer

Answer» Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
10720.

If m th term of an AP is 1/n and n th term is 1/m. Show that the sum of the mn terms is 1/2 (mn+1)

Answer» Given {tex}m^{th}term=\\frac{1}{n}{/tex}\xa0i.e. {tex}{/tex}\xa0{tex}a_m=\\frac{1}{n}{/tex}{tex}\\Rightarrow (a+(m-1)d)=\\frac{1}{n}{/tex} ......(1)and{tex}n^{th}term=\\frac{1}{m}{/tex}i.e. {tex}a_n=\\frac{1}{m}{/tex}{tex}\\Rightarrow a+(n-1)d=\\frac{1}{m}{/tex}.....(2)subtracting Eq(1) from\xa0Eq(2) , we get\xa0{tex}\\Rightarrow[ a+(n-1)d] -[(a+(m-1)d)] =\\frac{1}{m} - \\frac{1}{n}{/tex}{tex}\\Rightarrow a+(n-1)d -a+(m-1)d =\\frac{n - m}{mn} {/tex}{tex}\\Rightarrow nd- d -md+d =\\frac{n - m}{mn} {/tex}{tex}\\Rightarrow (n -m)d =\\frac{n - m}{mn} {/tex}{tex}d=\\frac{1}{mn}{/tex}\xa0and {tex}a=\\frac{1}{mn}{/tex}Now, sum of mn terms,{tex}{S_{mn}} = \\frac{{mn}}{2}[2a + (mn - 1)d]{/tex}{tex}\\Rightarrow[{S_{mn}} = \\frac{{mn}}{2}[\\frac{2}{{mn}} + (mn - 1)\\frac{1}{{mn}}]{/tex}={tex}\\frac{1}{2}(mn+1){/tex}\xa0Hence Proved.
10721.

Find the probability of getting 53 Sundays in (a) leap year , (b) a non leap year

Answer» (a)2/53 (b)1/53
10722.

Sin theta

Answer» Sin$=p/h
10723.

10_1000

Answer»
10724.

An

Answer»
10725.

Ram has 2 pens and 3eraser how many total thing has Ram?

Answer» 5
10726.

45

Answer»
10727.

Q. 3 If the sum of zeroes of quadratic polynomial 3x square - Kx +6 is 3, then fine the value of k

Answer» Solve it in proper manner..
9
Can you please solve it properly
8.48
10728.

Sec A +cosec A -tan A=???

Answer» Answer is sin A Cos A
10729.

What is 4q+1 and 4q

Answer» Let a = 4q + r, when r = 0, 1, 2 and 3{tex}\\therefore{/tex}Numbers are 4q, 4q + 1, 4q + 2 and 4q + 3{tex}( a ) ^ { 2 } = ( 4 q ) ^ { 2 } = 16 q ^ { 2 } = 4 ( 4 q ) ^ { 2 } = 4 m{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 1 ) ^ { 2 } = 16 q ^ { 2 } + 8 q + 1 = 4 \\left( 4 q ^ { 2 } + 2 q \\right) + 1 = 4 m + 1{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 2 ) ^ { 2 } = 16 q ^ { 2 } + 16 q + 4 = 4 \\left( 4 q ^ { 2 } + 4 q + 1 \\right) = 4 m{/tex}{tex}( a ) ^ { 2 } = ( 4 q + 3 ) ^ { 2 } = 16 q ^ { 2 } + 24 q + 9 = 4 \\left( 4 q ^ { 2 } + 6 q + 2 \\right) + 1 = 4 m + 1{/tex}{tex}\\therefore {/tex}\xa0the square of any +ve integer is of the form 4q or 4q + 1\xa0
10730.

Show that the points (1,1),(4,4),(4,8),(1,5) form a parallelogram

Answer»
10731.

Please send chapter wise marking system of new case syllabus 2018

Answer»
10732.

Find roots of equation

Answer»
10733.

In board which type of question ask?hard or easy..

Answer» They are easy but they will change the questions and answer will be the same.
Don\'t worry, it will be easy if you study
10734.

880 is in root than what comes out

Answer» 4√55
10735.

Find the 5th term the end of the A.P 17, 14, 11....., -40

Answer» Here, a= 17 and d=(14-17)=-3 Since, a+(n-1)d17+(5-1)×-3=17+(4)×-3=17-12=5Therefore,5th term is 5
10736.

Sec a +Tan a _1/Tan a _Sec a +1=Cos a /1+Sin a

Answer»
10737.

What are the composite number

Answer» Composite Number:A whole number that can be divided evenly by numbers other than 1 or itself. Example: 9 can be divided evenly by 3 (as well as 1 and 9), so 9 is a composite number. But 7 cannot be divided evenly (except by 1 and 7), so is NOT a composite number (it is a prime number).
10738.

solve 3x + y=3 and 9x-3y=9.

Answer» 3x - y = 39x - 3y = 9The given pair of linear equations is\xa03x - y = 3..............(1)9x - 3y = 9.............(2)From equation(1),\xa0y = 3x - 3...................(3)9x - 3(3x - 3) = 9{tex}\\Rightarrow{/tex}\xa09x - 9x + 9 = 9{tex}\\Rightarrow{/tex}\xa09 = 9which is true. Therefore, equation (1) and (2) have infinitely many solutions.
10739.

Find the root of the following quardtic equations by the method of completing square 2x²-7x+3=0

Answer» We have 2x2 - 7x + 3 = 0{tex}\\implies2( x^2 - {7 \\over 2}x + {3\\over 2}) = 0{/tex}{tex}\\implies\u200b\u200b x^2 - {7 \\over 2}x + {49 \\over 16} = {-3 \\over 2} +{ 49 \\over 16}{/tex} (Adding 49/16 to both sides){tex}\\implies x^2 -2 \\times x \\times {7 \\over 4} + ({7 \\over 4})^2 = {-24 +49 \\over 16}{/tex}{tex}\\implies (x-{7\\over4})^2 = {25 \\over 16}{/tex}{tex}\\implies x-{7\\over 4}= \\pm \\sqrt({25 \\over 16}){/tex}{tex}\\implies x={7\\over 4} \\pm {5 \\over 4}{/tex}{tex}\\implies x={7\\over 4} + {5 \\over 4}\\, and \\,x={7\\over 4} - {5 \\over 4}{/tex}{tex}\\implies x=3\\, and \\,{1\\over 2}{/tex}{tex}\\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\\over 2{/tex}.
10740.

What is Bpt teorm

Answer» It state that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points the other 2 sides are divided in the same ratio
10741.

Using Euclid division algorithm find hcf of135 and 225

Answer» On applying the division lemma to 225 and 135\xa0We get225=135×1+90135=90×1+4590=45×2+0Hence HCF(225,135)=45
10742.

If sinA+cosA=x then prove that sin6A+cos6A= 1+(x2-1)2/4

Answer»
10743.

Solve tsquare +underoot15 with quadatic formula

Answer»
10744.

find the value of p when the distance between the A(3,p) B(4,1)is underroot 10

Answer» We have P(3, a) and Q(4,1)Here,\xa0x1 = 3, y1 = ax2 = 4, y2 = 1PQ =\xa0{tex}\\sqrt { 10 }{/tex}{tex}P Q = \\sqrt { \\left( x _ { 2 } - x _ { 1 } \\right) ^ { 2 } + \\left( y _ { 2 } - y _ { 1 } \\right) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { ( 4 - 3 ) ^ { 2 } + ( 1 - a ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { ( 1 ) ^ { 2 } + ( 1 - a ) ^ { 2 } }{/tex}{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { 1 + 1 + a ^ { 2 } - 2 a }{/tex}\xa0{tex}\\Rightarrow \\quad \\sqrt { 10 } = \\sqrt { 2 + a ^ { 2 } - 2 a }{/tex}Squaring both sides{tex}\\Rightarrow ( \\sqrt { 10 } ) ^ { 2 } = \\left( \\sqrt { 2 + a ^ { 2 } - 2 a } \\right) ^ { 2 }{/tex}{tex}\\Rightarrow{/tex}\xa010 = 2 + a2 - 2a{tex}\\Rightarrow{/tex}\xa0{tex}a^2 - 2a + 2 - 10 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}a^2 - 2a - 8 =0{/tex}Splitting the middle term.{tex}\\Rightarrow{/tex}\xa0a2 - 4a + 2a - 8 = 0{tex}\\Rightarrow{/tex}\xa0{tex}a(a - 4) + 2 (a - 4)=0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(a - 4)(a + 2) = 0{/tex}{tex}\\Rightarrow{/tex}{tex}\xa0a = 4, a = -2{/tex}
10745.

Cosec31 -sec59ka man gyat kare

Answer» Cosec(90-59)-sec59=sec59-sec59=0
10746.

Sin A -Cos B ज्ञात कीजिए A+B और कोण A और B न्यून कोण है

Answer»
10747.

I want half yearly sample paper with solution.

Answer» samples
10748.

How can we find area of isosceles triangle. If two sides are 20cmand12cm.Area is 60cmsquare?

Answer» Yes
10749.

Wht is the syllabus of 10th class for SA 1??

Answer» ******* mam ne na likh wayo
10750.

Good morning friends next mera mathematics ka exam hai ?

Answer»