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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 36101. |
1245+7890 |
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Answer» 9135 9135 Kgp 9135 |
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| 36102. |
If 1-tan^2theta =30degree then write the value of sintheta + cos^2theta |
| Answer» Ans = 5/4 | |
| 36103. |
AD=4 cm bc=12 cm Bd=3 cm and find cot , sec, and cosec angle c =thetha |
| Answer» | |
| 36104. |
a3 =15,s10=125, find d and a10 |
| Answer» Here, a3 = 15S10 = 125We know thatan = a + (n - 1)d{tex} \\Rightarrow {/tex}\xa0a3 = a + (3 - 1)d{tex} \\Rightarrow {/tex}\xa0a3 = a + 2d{tex} \\Rightarrow {/tex}\xa015 = a + 2d{tex} \\Rightarrow {/tex}\xa0a + 2d = 15 ...... (1)Again, we know that{tex}{S_n} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex} \\Rightarrow {S_{10}} = \\frac{{10}}{2}\\left[ {2a + (10 - 1)d} \\right]{/tex}{tex} \\Rightarrow {/tex}\xa0S10 = 5(2a + 9d){tex} \\Rightarrow {/tex}\xa0125 = 5(2a + 9d){tex} \\Rightarrow {/tex}\xa025 = 2a + 9d{tex} \\Rightarrow {/tex}\xa02a + 9d = 25 ....... (2)Solving equation (1) and equation (2), we geta = 17d = -1Now an = a + (n - 1)d{tex} \\Rightarrow {/tex}\xa0a10 = a + (10 - 1)d{tex} \\Rightarrow {/tex}\xa0a10 = a + 9d{tex} \\Rightarrow {/tex}\xa0a10 = 17 + 9(-1){tex} \\Rightarrow {/tex}\xa0a10 = 17 - 9{tex} \\Rightarrow {/tex}\xa0a10 = 8 | |
| 36105. |
Can you proove 5=7 |
| Answer» How is it possible | |
| 36106. |
Html code forCa(OH)2 + CO2 - CaCO3 + H2O |
| Answer» | |
| 36107. |
How to solve a ap |
| Answer» By hand | |
| 36108. |
X +y=78X+y×y=144 |
| Answer» | |
| 36109. |
Supposé the altitudes of à triangle are 10,12,and 15. What is its semi-perimeter. |
| Answer» 18.5 | |
| 36110. |
ax2+bx+c=0 |
| Answer» It us a general form of quadratic equation | |
| 36111. |
What is blue print of class 10cbse |
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Answer» Blue print bhi hota h human being me... Really I don\'t know DNA is blue print in humqn body |
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| 36112. |
If tan = 1, find 2sin cos |
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Answer» A simple and quick methos------------------Tan 45°=1Therefore theta = 45°Therefore 2sin.cos= sin(2theta)= sin (2×45)°=sin 90°=【1】 It will be equal to 1 as tan45=1and 2sincos also =1 2 |
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| 36113. |
Prove that. tan/1-cot+cot/1-tan=(1+sec cosec) |
| Answer» LHS =\xa0{tex}\\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta }{/tex}{tex}= \\frac { \\frac { \\sin \\theta } { \\cos \\theta } } { 1 - \\frac { \\cos \\theta } { \\sin \\theta } } + \\frac { \\frac { \\cos \\theta } { \\sin \\theta } } { 1 - \\frac { \\sin \\theta } { \\cos \\theta } }{/tex}\xa0{tex}\\left[ \\because \\tan \\theta = \\frac { \\sin \\theta } { \\cos \\theta } , \\cot \\theta = \\frac { \\cos \\theta } { \\sin \\theta } \\right]{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } + \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\cos \\theta - \\sin \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } - \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 3 } \\theta - \\cos ^ { 3 } \\theta } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}\xa0{tex}= \\frac { ( \\sin \\theta - \\cos \\theta ) \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta + \\sin \\theta \\cos \\theta \\right) } { ( \\sin \\theta - \\cos \\theta ) \\sin \\theta \\cos \\theta }{/tex}\xa0[ a3\xa0- b3\xa0= (a - b)(a2\xa0+ ab + b2) ]{tex}= \\frac { 1 + \\sin \\theta \\cos \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\frac { 1 } { \\sin \\theta \\cos \\theta } + 1 = 1 + \\sec \\theta cosec \\theta{/tex}\xa0= RHStherefore, RHS = LHS | |
| 36114. |
Introduction of trigonometry |
| Answer» Relationship between angles and side of a right angle triangle | |
| 36115. |
A table-top measures 2m 25cm by 1m 50 cm what is the perimiter of the table-top |
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Answer» 7m 50cm 750 |
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| 36116. |
one arkmeans |
| Answer» | |
| 36117. |
If tanA=cotB,prove that A+B=90 |
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Answer» Tan a = cot bTan (90-a)=cot bCot( 90-A)= cot B90-A=B90=B+ANow; A+B=90 Hence proved Tan a=cot bCot (90-a) = cot b90-a=b90=a+b A+b=90 It is ask in1-2 no. In exam |
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| 36118. |
if a cosec A = p and b cotA = q prove that p square /a square -q square / b square=1 |
| Answer» | |
| 36119. |
Cos A - sin A +1 /cos A +sin A -1 = cosec A+cotA, using the identity cosec square A= 1+cot square A |
| Answer» | |
| 36120. |
Prove that √7is a irrational number |
| Answer» | |
| 36121. |
2+2 is 4-1 that\'s 3 Quick maffs! |
| Answer» | |
| 36122. |
10\\x+y+2/x_y |
| Answer» | |
| 36123. |
Way to learn trigonometry |
| Answer» Learn only sin cos and tan for sin = opposite side by hypotensecos = adjecent side by hypotensetan = sinby cosAnd other three by reversing the aboves | |
| 36124. |
can i get evergeeen solution here |
| Answer» Why u get evergreen solution here | |
| 36125. |
Solve the quadratic equation:-16/x -1=15/x+1 |
| Answer» We have {tex}\\frac{{16}}{x} - 1 = \\frac{{15}}{{x + 1}}{/tex}{tex} \\Rightarrow \\frac { 16 - x } { x } = \\frac { 15 } { x + 1 }{/tex}Cross multiply,{tex}\\Rightarrow{/tex}\xa0{tex}(16-x)(x+1)=15x{/tex}{tex} \\Rightarrow{/tex}\xa0{tex}16x+16-x^2-x=15x{/tex}{tex} \\Rightarrow{/tex} 15x - 16x - 16 + x2 + x = 0{tex} \\Rightarrow{/tex} x2 - 16 = 0{tex} \\Rightarrow{/tex} x2 = 16{tex} \\therefore{/tex} x = {tex} \\pm{/tex}4 | |
| 36126. |
How can I make my ap ch5 strong |
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Answer» My ligin\' ma balls! Solving all question freely and shared your doubt to all of them and do numericals by s.chand and r.s aggrawal |
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| 36127. |
Find the value of "a" so that the point (3, a) lie on the Represented by 2x - 3y = 5 |
| Answer» Since the point (3, a) lies on the line 2x - 3y = 5, we have2\xa0{tex}\\times{/tex}\xa03 - 3\xa0{tex}\\times{/tex}\xa0a = 5{tex}\\Rightarrow{/tex}\xa06 - 3a = 5{tex}\\Rightarrow{/tex}\xa03a = 1{tex}\\Rightarrow{/tex}\xa0{tex}a = \\frac { 1 } { 3 }{/tex} | |
| 36128. |
For what value of p, are 2p-1, 7 and 3p three consecutive term of A.P? |
| Answer» | |
| 36129. |
Is it required to for class 10 board exams from rd sharma?Ncert is sufficient for class 10 |
| Answer» It is sufficient if do everything perfectly and theory perfectly as it gives us an idea to solve every question | |
| 36130. |
Class 10 math |
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Answer» Its theorm :) Tum kya puchna chahte ho maths se?? koi ques.,koi therom........... ..? |
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| 36131. |
trignometry of table kase banate h |
| Answer» | |
| 36132. |
State and prove pythagorous theorem |
| Answer» a²+b²=c² | |
| 36133. |
Show that one and only one of n;n+4;n+2 is divisible by 3 |
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Answer» Take n =3qSo first n is divisible by 3Both n+4;n+2is also divided by 3 but it leaves reminder and n does not leave any reminder so n is divisible by 3 N |
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| 36134. |
{1+an²A\\1+cot²A}= {1- tanA\\cotA}²= tan ²A prove following identities |
| Answer» {tex}= \\frac { 1 + \\tan ^ { 2 } A } { 1 + \\cot ^ { 2 } A } = \\frac { 1 + \\tan ^ { 2 } A } { 1 + \\frac { 1 } { \\tan ^ { 2 } A } } \\cdot \\because \\cot A = \\frac { 1 } { \\tan A }{/tex}{tex}= \\frac { 1 + \\tan ^ { 2 } A } { \\frac { \\tan ^ { 2 } A + 1 } { \\tan ^ { 2 } A } } = \\tan ^ { 2 } A \\ldots \\ldots ( 1 ){/tex}{tex}\\left( \\frac { 1 - \\tan A } { 1 - \\cot A } \\right) ^ { 2 } = \\left( \\frac { 1 - \\tan A } { 1 - \\frac { 1 } { \\tan A } } \\right) ^ { 2 }{/tex}{tex}= \\left\\{ \\frac { 1 - \\tan A } { \\left( \\frac { \\tan A - 1 } { \\tan A } \\right) } \\right\\} ^ { 2 } = ( - \\tan A ) ^ { 2 } = \\tan ^ { 2 } A{/tex} ....... (2)(1) and (2) taken together given the result. | |
| 36135. |
If 3sin0+5cos0=5 prove that 5sin0-3cos0=+3 and -3 |
| Answer» R.s.agarwal ka page no.213 | |
| 36136. |
xSin π/6.〖Cos〗^2 π/4=(Tan π/4.Sec π/3.〖Cos〗^2 π/6)/(〖Cosec〗^(2 ) π/4.Sec π/6)find X |
| Answer» Pls provide for above question | |
| 36137. |
Measurements of weight |
| Answer» 1 pound (lb) = 16 ounces (oz) = 0.454 kilogram (kg) | |
| 36138. |
Solve by competing square method. (5x)5x+11x+9=0 |
| Answer» Ncert question or rd sharma | |
| 36139. |
4x^2+x-4 |
| Answer» | |
| 36140. |
Write all the other trigonometric ratio of Angle A in the terms of sec a |
| Answer» Cos A=1/sec ASin A=√1-cos sq. A=√1-1/sec sq.A=√sec sq.A-1/sec sq.ATan A=√sec sq.A-1Cot A=1/tan A=1/√sec sq.A-1Cosec A=1/sin A=1/√sec sq.A-1/sec A | |
| 36141. |
introduction for triangles |
| Answer» Introduction | |
| 36142. |
SinX +Sin3X +Sin5X+Sin7X=? |
| Answer» | |
| 36143. |
Prove that 0+0=1 |
| Answer» (0+0)never be 1 | |
| 36144. |
the sum of n positive integer |
| Answer» n/2×(a+l) and n/2×(2a+(n-1)×d) | |
| 36145. |
Solve for x- 1/a+b+x= 1/a+1/b+1/X |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + x ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { x }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + x ) } - \\frac { 1 } { x } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { x - ( a + b + x ) } { x ( a + b + x ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { x ( a + b + x ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { x ( a + b + x ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0x(a + b + x) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0x2 + ax + bx + ab = 0{tex}\\Rightarrow{/tex}\xa0x(x +a) + b(x +a) = 0{tex}\\Rightarrow{/tex}\xa0(x\xa0+ a) (x + b) = 0{tex}\\Rightarrow{/tex}\xa0x + a = 0 or x + b = 0{tex}\\Rightarrow{/tex}\xa0x = -a or x = -b.Therefore, -a and -b\xa0are the roots of the equation. | |
| 36146. |
If two vertices of equaliteral triangle are (0,0) and(3,root3). Find the third vertices |
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Answer» Thanx First of all let us assume that the cordinates of third side is (x,y) and we know that all sides of triangle are equal so , we use distance formula and find the distance of one side from the two coordinates and then find the distance of another two sides in terms of x and y and then corelate the equation |
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| 36147. |
Can anyone will teach me trigonometry plz? |
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Answer» Pakistan bhukha pyasa, pbp. Hindustan Hara bhara PBP-Papa Bear Piyoge ;HHB- Ha Ha Beta PBP Pandit Badri Prsad1)--------= ------------------------------- HHB Har Har Bhole2)sinA=P/H CosA=B/H tanA=P/B Cosec=H/P Cot=H/B Sec=B/P Yes |
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| 36148. |
Prove that root 6 is an irrational numbers |
| Answer» First assume that √6 is arational no. Then , √6= p by q form ,where p & q are integers ,q is not equals to zero and p&q are co-primes .Squaring both the sides 6= p sq. by q sq. = p sq.= 6q sq. -[1]i.e. p sq. is divisible by 6. So, p is also divisible by 6.Then , we can write p= 6rsquaring both the sides p sq.=12r sq.2 q sq= 4r sq. -[from equ. (1)]q sq. =4r sq. by 2 => q sq. =2r sq. -(2)i.e. q sq. Is divisible by 2So, q is also divisible by 6. Since, we can find p&q have a common factor between them (i.e. 2)In our contradicts & assumption √3 is not a rational number. It is an irrational no. | |
| 36149. |
2÷10=2Prove it |
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Answer» When 2 is divided by 10 then it gives us quotient 0.2 totally wrong ques???? The ques. is wrong bread.... |
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| 36150. |
The product of two numbers is 396 x 576 and their LCM is 6336.Find their HCF |
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Answer» Such a easy question 36 |
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