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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.
| 11901. |
X=-2 and x=-1/5 are solutions find p and r for equations 5x square +px +r=0 |
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| 11902. |
Triangle ABC is similar to triangle ADE if AE =2cm. EC=3cm and DE=1.6cm then find BC |
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| 11903. |
the arithmetic means of the following data is 25, find the value of k |
| Answer» The frequency table of the given data is as given below:\tValue (xi)141516182025kFrequency (fi)1313331\tIt is given that the mode of the given date is 25. So, it must have the maximum frequency. That is possible only when k\xa0=25.Hence, k\xa0= 25. | |
| 11904. |
Prove that:Sec8a-1/sec4a-1 = tan8a/tan2a |
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| 11905. |
Calculate the modal income |
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| 11906. |
(x-3)(2x+1)=x(x+5) |
| Answer» 5+✓57/2,5-✓57/2 | |
| 11907. |
How i get 80 above percent in class 10th |
| Answer» Do NCERT 1st of all, Each question & Example....Once done, then pick up the last 5 year question papers which u may get in n number of books / online websites.STUDY HARD... AND ATTEMPT SMARTLY without cutting & mistakes. ... ALL THE BEST\xa0 | |
| 11908. |
152x-378y=-74 and -378x+152y=-604 |
| Answer» The given equations are{tex}152x - 378y = -74{/tex} ...(i){tex}-378x\xa0+ 152y = -604{/tex}. ... (ii)Clearly, the coefficients of x\xa0and y in one equation are interchanged in the other.Adding (i) and (ii), we get{tex}(152-378)x\xa0+ (-378 +152) y = -(74 + 604){/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(-226)x\xa0+ (-226)y = -678{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}(-226)(x + y) = -678{/tex}{tex}\\Rightarrow \\quad ( x + y ) = \\frac { - 678 } { - 226 } \\Rightarrow x + y = 3{/tex}. ...(iii)Subtracting (ii) from (i), we get{tex}(152 + 378) x + (-378 - 152)y = (-74 + 604){/tex}{tex}\\Rightarrow{/tex}530x\xa0- 530y = 530{tex}\\Rightarrow{/tex}\xa0530(x - y) = 530 {tex}\\Rightarrow{/tex}\xa0x\xa0- y = 1. ... (iv)Adding (iii) and (iv), we get 2x = 4 {tex}\\Rightarrow{/tex}\xa0x = 2.Subtracting (iv) from (iii), we get2y = 2 {tex}\\Rightarrow{/tex} y = 1.Hence, x = 2 and y = 1 | |
| 11909. |
Statement of (AAA - similarty ) |
| Answer» AAA Similarity Criterion: If in two triangles, the corresponding angles are equal, then their corresponding sides are proportional and hence the triangles are similar. | |
| 11910. |
4+6 |
| Answer» 10 | |
| 11911. |
1/(x-3)-1/(x+5)=1/6 |
| Answer» X=-9 or x=7 | |
| 11912. |
Show that the polynomial f(x)= x to the power 4 + 4 x square + 6 has no zero |
| Answer» Given, f(x) = x4 + 4x2 + 6 = (x2)2 + 4x2 + 6Put x2 = aHence x4 + 4x2 + 6 = (x2)2 + 4x2 + 6 = a2 + 4a + 6To find the zero of the above polynomial, we take a2 + 4a + 6 = 0 We know D=b2-4ac .D=4^2-4*1*6 =16-24= -8 Here D<0 so it has no real root. | |
| 11913. |
If cosec theta/ sin theta - xtan^theta =1 |
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| 11914. |
Prove that 1-sin theta / 1+sin theta = 1-sin theta / cos theta |
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| 11915. |
What is the theorems in chapter circles |
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| 11916. |
Find the odd number 35752178 |
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| 11917. |
Where x plus x by 1 equal 3 .solve by factors methods |
| Answer» x + x\xa0= 32 x = 3 => x = 3/2\xa0(Answer)\xa0 | |
| 11918. |
?Trigonometric identities ??? |
| Answer» Sin^2+cos^2=11+tan^2=sec^21+cosec^2=cot^2 | |
| 11919. |
(x+1) (x-2) |
| Answer» X=2X=-1 | |
| 11920. |
Prove the BPT theorem |
| Answer» The theorem is if a line drawn parallel to one side of a triangle interest the other two side at a distinct point either it divdes in same ratio | |
| 11921. |
Does question papers are come from board in September exam |
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Answer» Paper was tough Yes |
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| 11922. |
Using Euclid division lemma find HTC of 56 , 96 and 404 |
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| 11923. |
x+3×2x-5=32 |
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Answer» x + 3 ×\xa02x - 5 = 32Applyng BODMAS, x + 6x - 5 = 32 => 7x - 5 = 32 => 7x = 37 => x = 37/7 (Answer)\xa0\xa0 I think roots are not realX=1+√377/4X=1-√377/4 |
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| 11924. |
How many side are there in treangle |
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Answer» 3 sides 3 |
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| 11925. |
In an acute angled triangle ABC,if sin (A+B-C)= 1/2 and cos (B+C-A)= 1/root 2 |
| Answer» Sin (A+B-C) = 1/2 => Sin (A+B-C) = Sin 300Cos (B+C-A) = 1/√2 => Cos (B+C-A) = Cos 450=> A + B - C = 300 ........ (Eq. 1)AND, B + C - A = 450 ........ (Eq. 2)ALSO, A + B + C = 1800 ........ (Eq. 3)Adding Eq. 1 & 2, we get: 2 B = 750 => B = 37.50Subtracting eq. 2\xa0from eq. 3, we get: 2 A = 1350\xa0=> A = 67.50Using value of A & B, in Eq. 3, we get C = 750\xa0\xa0\xa0\xa0 | |
| 11926. |
Show that 15 power n can never end up with zero |
| Answer» 15n\xa0= 15\xa0* 15 * 15 * 15 *15 *15 *............... nLet n = 1 => 15For n = 2,\xa0=> 225For n = 3, => 3375\xa0Plz note that for every value of n, the last digit ends with 5 everytime, thus the last digit will never be other than 5.\xa0 | |
| 11927. |
What is value of sin30 |
| Answer» 1/2 | |
| 11928. |
how we can easily proof trignometric eq. |
| Answer» By changing LHS and RHS in sin ,cos If eq. already in sin,cos then divide numerator and Denominator by sin | |
| 11929. |
7sin^2A+3cos^2A=4 show that tanA=1/root3 |
| Answer» {tex}7sin^2A+3cos^2A=4{/tex}{tex}=> 4sin^2A+ 3sin^2A+3cos^2A=4{/tex}{tex}=> 4sin^2A+ 3(sin^2A+cos^2A)=4{/tex}{tex}=> 4sin^2A+ 3=4{/tex}{tex}=> 4sin^2A = 1 {/tex}{tex}=> sin^2A = {1\\over 4} {/tex}{tex}=> sinA = {1\\over 2}{/tex}{tex}=> sinA = sin 30^o{/tex}{tex}=> A = 30^o{/tex}So\xa0{tex}tan A= tan 30^o = {1\\over \\sqrt 3}{/tex} | |
| 11930. |
Trick of AP |
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| 11931. |
Find the sum of all the 11 terms of an AP whose middle most term is 30 |
| Answer» given n =11middle term = (a1 + a11)/2 = 30a1 + a11 = 60 .......... iSn = n/2(a1 + a11) ......... ii = 11/2 (60) ... from i = 11 x 30 = 330 | |
| 11932. |
Find the sum of all the 11 terms of an Ap whose middle most term is 30. |
| Answer» Let a be the first term and d be the common difference of the given A.P.Clearly, in\xa0an A.P. consisting of 11 terms,\xa0{tex} \\left( \\frac { 11 + 1 } { 2 } \\right) ^ { t h }{/tex}\xa0i.e. 6th term is the middle term.{tex}\\text{ it is given that the middle term =30}{/tex} So a+5d=30 .....(1){tex}S_{11}=\\frac{11}{2}(2a+10d){/tex}\xa0= 11(a+5d) But a+5d=30 from (1)Hence S\u200b\u200b\u200b\u200b\u200b\u200b11\xa0= 11 × 30= 330 | |
| 11933. |
What is the full form of A.T.P.ANS |
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Answer» adenosine triphosphate it is the energy currency Adenosine Triphosphate |
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| 11934. |
Prove Pythagoras theorem ................. |
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| 11935. |
cicke formulas |
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| 11936. |
Sin2a+sin2b=1 |
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| 11937. |
Hamare 10th k maths ka exam ncert m see hi hoga kya |
| Answer» My was from ncert | |
| 11938. |
Prove that 1-cosA+sinA/sinA+cosA-1=1+sinA/cosA |
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| 11939. |
Application of tri |
| Answer» Use to measure distances | |
| 11940. |
TanA/cotA+tanB/cotB =tanAtanB |
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| 11941. |
Median divide in two equal ratio but why |
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| 11942. |
state fundamental theorem of arithmetic |
| Answer» it states that any positive integer greater than 1 is either a prime number or a product of prime numberse.g.2 is a prime no3 is a prime no4 = 2x2 = 225 is prime6 = 2x37 is prime8 = 2x2x2 = 23 and so on | |
| 11943. |
Sin2A+ Cos2A |
| Answer» Sin2A+Cos2A=1 hota Hai. | |
| 11944. |
Prove that 0/0=2 ?? |
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| 11945. |
what is alternate segment theorem |
| Answer» It states that an angle between tangent and a chord of a circle through the point of contact is equal to the angle in the alternate segment | |
| 11946. |
Find the quadratic polynomial whose zeros are 7+√3 and 7+√3 |
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| 11947. |
Find zeros of polynomials |
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| 11948. |
ABC is a triangle angle A = Angle B find angle C |
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| 11949. |
2*5x+8y |
| Answer» 5x + 4y | |
| 11950. |
Find roots of equation :ax+bx+c |
| Answer» roots of equation ax2+bx+c=0 where a , b , c are real numbers are given by{tex}x = {-b \\pm \\sqrt{b^2-4ac} \\over 2a}{/tex} | |