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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.
| 26951. |
Value of cot45 degree |
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Answer» 1 1 |
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| 26952. |
If theeta is 90° |
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Answer» Bhai poraa ques.likho ye complete question nahi hai No Is zero is prime number |
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| 26953. |
Please answer my question... |
| Answer» Bhai aapka question kya hai please tell me☺? | |
| 26954. |
(1+cotA+tanA)(sinA - cosA) |
| Answer» L.H.S\xa0= (1 + cotA + tanA) (sinA - cosA)= sinA - cosA + cotA sinA - cotA cosA + sinA tanA - tanA cosA{tex}= \\sin A - \\cos A + \\frac{{\\cos A}}{{\\sin A}} \\times \\sin A - \\cot A\\cos A + \\sin A\\;\\tan A - \\frac{{\\sin A}}{{\\cos A}} \\times \\cos A{/tex}= sinA - cosA + cosA - cotA cosA + sinA tanA - sinA= sinA tanA - cotA cosA........(1)Now taking ;{tex}\\quad \\frac{{\\sec A}}{{\\cos e{c^2}A}} - \\frac{{\\cos ecA}}{{{{\\sec }^2}A}}{/tex}{tex} = \\frac{{\\frac{1}{{\\cos A}}}}{{\\frac{1}{{{{\\sin }^2}A}}}} - \\frac{{\\frac{1}{{\\sin A}}}}{{\\frac{1}{{{{\\cos }^2}A}}}}{/tex}{tex} = \\frac{{{{\\sin }^2}A}}{{\\cos A}} - \\frac{{{{\\cos }^2}A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\frac{{\\sin A}}{{\\cos A}} - \\cos A \\times \\frac{{\\cos A}}{{\\sin A}}{/tex}{tex} = \\sin A \\times \\tan A - \\cos A \\times \\cot A{/tex}.......(2)From (1) & (2),(1 + cotA + tanA) (sinA - cosA) =\xa0{tex}\\frac { \\sec A } { cosec ^ { 2 } A } - \\frac { cosec A } { \\sec ^ { 2 } A }{/tex}\xa0= sinA.tanA - cosA.cotA\xa0Hence, Proved. | |
| 26955. |
Find the value of k for which the given equation have real and equal roots :x²-2(k+1)x+k² |
| Answer» {tex}k = \\frac { - 1 } { 2 }{/tex}\xa0 | |
| 26956. |
Find a quadratic polynomial whose zero are 4+√5 and 4-√5 |
| Answer» Let α=4\xa0-\xa0{tex}\\sqrt 5{/tex}and\xa0β = 4 +\xa0{tex}\\sqrt 5{/tex}Sum of zeroes,α+β = 4\xa0-\xa0{tex}\\sqrt 5{/tex}\xa0+ 4\xa0+\xa0{tex}\\sqrt 5{/tex}\xa0= 8αβ = (4\xa0-\xa0{tex}\\sqrt 5{/tex})(4 +\xa0{tex}\\sqrt 5{/tex}) = 16 - 5 = 11Quadratic polynomialp(x) = x2\xa0- (α+β)x +αβ= x2\xa0-\xa08x + 11 | |
| 26957. |
If d is HCF of 45 and 27 find x and y satisfying d=45x+28y |
| Answer» SIMILAR QUESTION:\xa0If d is the hcf of 45 and 27,find x and y satisfying d= 27x+45yThe numbers are :-45 and 27Lets find their HCFBY Euclid\'s division Lemma :-a = bq + r45 = 27 x 1 + 1827 = 18 x 1 + 918 = 9 x 2 + 0HCF = d = 9Given:-d = 27x+45y9 = 27x + 45y9 = 27 - 18 x 118 = 45 - 27 x 1\xa0Therefore ,9 = 27 - [ 45 x 27 x 1 ] x 1= 27 x 2 - 45\xa09 = 27 x 2 + 45 x [ -1 ]x = 2y = -1 | |
| 26958. |
Class 10 RS AGGARWAL new additionPg 59 question no 14 answer |
| Answer» 63737 | |
| 26959. |
Exercise 6.5 question no 2 |
| Answer» PQR is a triangle right angled at P and M is a point on QR such that PM\xa0\xa0QR. Show that PM2\xa0= QM x MR.Ans. Given: PQR is a triangle right angles at P and PMQRTo Prove: PM2\xa0= QM.MRProof: Since PMQR\xa0QMPPMR\xa0\xa0 | |
| 26960. |
If Sin^2 A = SinA * CosA than A is equal to |
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Answer» Aaya mahatma ji ka answer aya? 45° |
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| 26961. |
4u 4u square + 8 you |
| Answer» 4u2\xa0+ 8u = 04u(u + 2) = 0u = 0 or u = -2 | |
| 26962. |
Show the every positive odd integer is the form 4m+1 and 4m+3 where m is form of some integers |
| Answer» Step-by-step explanation:Taking q as some integer.Let a be the positive integer.And, b = 4 .Then by Euclid\'s division lemma,We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .°•° Then, possible values of r is 0, 1, 2, 3 .Taking r = 0 .a = 4q .\xa0Taking r = 1 .\xa0a = 4q + 1 .Taking r = 2a = 4q + 2 .Taking r = 3 .a = 4q + 3 .But a is an odd positive integer, so a can\'t be 4q , or 4q + 2 [ As these are even ] .•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .Hence , it is solved | |
| 26963. |
Show that n2-1 is divisible by8,if n is an odd positive integers |
| Answer» Are bhai n=4m+1or4m+2leke dekho ban jayega. | |
| 26964. |
Evaluate cotA(90°-©)/sin©+cotA40°/tanA50°-cos2 20°+cos2+70° |
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| 26965. |
6 chapter |
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| 26966. |
For any positive integer n, proved that n×n×n_n is divisible by 6 |
| Answer» n3\xa0- n = n (n2\xa0- 1) = n (n - 1) (n + 1)\xa0Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.⇒ n (n – 1) (n + 1) is divisible by 3.\xa0Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.∴ n = 2q or 2q + 1, where q is some integer.If n = 2q, then n is divisible by 2.If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.⇒ n (n – 1) (n + 1) is divisible by 2.Since, n (n – 1) (n + 1) is divisible by 2 and 3.∴ n (n-1) (n+1) = n3\xa0- n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)\xa0\xa0 | |
| 26967. |
4 ko 4 baar use krke ans 20 lao. ( +,-,/,X) |
| Answer» | |
| 26968. |
Chapter-3. Exercise-3.3. Question 5 |
| Answer» You can check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 26969. |
Solve for x-1/1 +x-3/x-4=10/3; |
| Answer» X=2√2/3 | |
| 26970. |
√3x - √2 -√5 |
| Answer» | |
| 26971. |
Is R d sharma mathematics helpful for 10 th board exam |
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Answer» ?hmm Ryt chetna ji ? Welcome ji? Thanks chetna Yup....but firstly do ur NCERT if it\'s clear then u can move to RD |
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| 26972. |
Anybody is there |
| Answer» Yess yrr | |
| 26973. |
Hello dosto?? |
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Answer» Hi Hlow |
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| 26974. |
2÷1÷2÷3 |
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Answer» 3 0.333333333333333333Correct answer ☺️ |
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| 26975. |
If one zeroes of the quadratic polynomial (k-1) x^2+kx+1is -3, then find the value of k |
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| 26976. |
Koi bta skta h kya ki tumhare school me ncert book chl rhi hh ya RD sharma book math ki |
| Answer» Bro RD sharma ka maths S-Chand ke biology, chemistry, physics K.S Randhawa ke SST | |
| 26977. |
Find the values of A and B so that x cube + x square + bx - 6 is divisible by x square - 4 x + 3. |
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| 26978. |
Chlo m chli ...byyy ...byyyy .... See u later ....tc ..?? |
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Answer» ??????? Bbye..mai bhi ja rhi hun..gd nyt..tke care bestie??? |
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| 26979. |
Notes. Of chapter 14 |
| Answer» Source of enargy | |
| 26980. |
(Sin theta +sec theta)2 +(cosec theta + cos theta)2= (1+ sec theta + cosec theta)2 |
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Answer» Plz answer it Ashish just shut your mouth Jisko aata ho usne v to nhi kiya Runi yadav r u over smart.Ni aata to msg kyu kiya Sry nhi ataNt main dekh lo??? Plzzz... answer it |
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| 26981. |
9xx-9(a+b)x+(2aa+5ab+2bb)=0 |
| Answer» D = b2 - 4ac{tex}= ( - 9 ( a + b ) ) ^ { 2 } - 4 \\times 9 \\times \\left( 2 a ^ { 2 } + 5 a b + 2 a b ^ { 2 } \\right){/tex}= 81(a + b)2 - 36(2a2 + 5ab + 2b2)= 9[9(a2 + b2 + 2ab - 8a2 - 20ab - 8b2)]= 9[a2 + b2 - 2ab]= 9(a - b)2{tex}x = \\frac { - b \\pm \\sqrt { D } } { 2 a } = \\frac { 9 ( a + b ) \\pm \\sqrt { 9 ( a - b ) ^ { 2 } } } { 2 \\times 9 }{/tex}{tex}{ \\Rightarrow x = 3 \\frac { [ 3 ( a + b ) \\pm ( a - b ) ] } { 2 \\times 9 } }{/tex}{tex}{ \\Rightarrow x = \\frac { ( 3 a + 3 b ) \\pm ( a - b ) } { 6 } }{/tex}{tex}\\Rightarrow x = \\frac { 3 a + 3 b + a - b } { 6 } \\text { or } x = \\frac { 3 a + 3 b - a + b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 4 a + 2 b } { 6 } \\text { or } x = \\frac { 2 a + 4 b } { 6 }{/tex}{tex}\\Rightarrow x = \\frac { 2 a + b } { 3 } \\text { or } x = \\frac { a + 2 b } { 3 }{/tex} | |
| 26982. |
Two no. Are in the ratio 9:5 if there lcm is 900 then find hcf |
| Answer» | |
| 26983. |
Solve.2x+1/4-2x-3/6=7/3 |
| Answer» answer | |
| 26984. |
3×4 |
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Answer» 12 12 12 12 |
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| 26985. |
Hi.....sets. Kr liye kisine |
| Answer» Maine commerce liya hai | |
| 26986. |
points makes straight line or point is a part of straight line?? |
| Answer» Point is a part of straight line | |
| 26987. |
Find the HCF of 963 and 657 using euclids division lemma |
| Answer» Given numbers are 657 and 963 .Here, 657 < 963\xa0By using Euclid\'s Division algorithmm , we get963 = (657 × 1) + 306Here , remainder = 306 .So, On taking 657 as new dividend and 306 as the new divisor and then apply Euclid\'s Division lemma, we get657 = (306 × 2) + 45Here, remainder = 45\xa0So, On taking 306\xa0as new dividend and 45\xa0as the new divisor and then apply Euclid\'s Division lemma, we get306 = (45 × 6) + 36Here, remainder = 36So, On taking 45\xa0as new dividend and 36\xa0as the new divisor and then apply Euclid\'s Division lemma, we get45 = (36 × 1) + 9Here, remainder = 9So, On taking 36\xa0as new dividend and 9\xa0as the new divisor and then apply Euclid\'s Division lemma, we get36 = (9 × 4) + 0Here , remainder = 0 and last divisor is 9.\xa0Hence, HCF of 657 and 963 = 9. | |
| 26988. |
X=7+4√3 , find √x + 1/√x |
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Answer» Hello Hlo ? kaise ho? Hmm ? Just for tympass Wah... Abhi bh maths |
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| 26989. |
Hlo...hiiii |
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Answer» Hii Hiii aman .... Hlooo riya?? |
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| 26990. |
Factorise 2x square - 11 x + 15 |
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Answer» 2x_5 or x_3 3, 2.5 |
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| 26991. |
1/2-1/2-1/2-x |
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Answer» 1(2-x)/2-1(2-1)=2-x/9 1(2-x)/2-1(2-1)2-x/9 |
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| 26992. |
I x=3 then x3 -7×6=0 |
| Answer» Prove that the following are irrational1 1/√2Sir plz solve this question | |
| 26993. |
What are the values of x and y in the linear equation 2x-2y=1 |
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| 26994. |
Hlo ..... |
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Answer» Hi Hlo 789 Hi hi ....456... Hlo..hlo.1,2,3? |
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| 26995. |
Any one can help me in question |
| Answer» Ya | |
| 26996. |
Short and simple defination of probability |
| Answer» Maybe or maybe not | |
| 26997. |
(1)2x2-5root2+4=0. (2) 6x2+7x-10=0 by the method of completing the square |
| Answer» | |
| 26998. |
Hiii...anyone on?? |
| Answer» Yess i m | |
| 26999. |
Find polynomial whose zero is 2root3 and 3root3 |
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Answer» This is not answer x2 - (5root3 )x + 18 |
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| 27000. |
Halo...halo????? |
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Answer» Hii bestie❤ Mst. ..aap btao.. Hola.... Kaise ho bestie |
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