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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.
| 2901. |
1/(cos(theta)**2 %2B 1) %2B 1/(sin(theta)**2 %2B 1) %2B 1/(sec(theta)**2 %2B 1) %2B 1/(cosec**2*theta %2B 1)=2 |
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| 2902. |
Among two supplementary angles the measure of the larger angle is 44° more thanthe measure of the smaller. Find their measures. |
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| 2903. |
Among the two supplementary angles, themeasure of the larger angle is 36° more thanthe measure of smaller. Find their measures |
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| 2904. |
23. Find the roots of equation 5x2 - 6x 2 0 by.the method of completing the square. |
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Answer» 5x^2 -6x -2 = 0 x^2 -(6/5)x -(2/5) = 0 (x)^2 - [2.(3/5).x] + (3/5)^2 - (3/5)^2 -(2/5) = 0 [x-(3/5)]^2 = (2/5)+(9/25) [x-(3/5)]^2 = (19/25) x-(3/5) = √19/5 x = (3/5) ± √19/5 Roots of the equation are x = [3+(√19)]/5 & x = [3-(√19)]/5 please like the solution 👍 ✔️ |
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| 2905. |
Find the roots of the equation 5x2-6x-20 by the method of completingthe square |
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| 2906. |
Find the root of the equation 5x2-6x-2-0 by the method ofcompleting the square. |
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| 2907. |
äš=0, 112Example 8: Find the roots of the equation 5x2-6x-2 0 by the method of completingthe square. |
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Answer» Given equation5x^2 -6x -2 = 0 x^2 -(6/5)x -(2/5) = 0 (x)^2 - [2*(3/5)*x] + (3/5)^2 - (3/5)^2 -(2/5) = 0 [x-(3/5)]^2 = (2/5)+(9/25) [x-(3/5)]^2 = (19/25) x-(3/5) = √19/5 x = (3/5) ± √19/5 Roots of the equation arex = [3+(√19)]/5, x = [3-(√19)]/5 |
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| 2908. |
Example 8: Find the roots of the equation 5x2-6x-2 0 by the method of completingthe square. |
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Answer» 5x^2 -6x -2 = 0 x^2 -(6/5)x -(2/5) = 0 (x)^2 - [2.(3/5).x] + (3/5)^2 - (3/5)^2 -(2/5) = 0 [x-(3/5)]^2 = (2/5)+(9/25) [x-(3/5)]^2 = (19/25) x-(3/5) = √19/5 x = (3/5) ± √19/5 Roots of the equation are x = [3+(√19)]/5 & x = [3-(√19)]/5 |
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| 2909. |
0 Example 8: Find the roots of the equation 5x2 -6x-2-0 by the method of completingthe square. |
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| 2910. |
Example 8: Find the roots of the equation 5x2-6x-2-0 by the method of completingthe square. |
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Answer» 5x^2 -6x -2 = 0 x^2 -(6/5)x -(2/5) = 0 (x)^2 - [2.(3/5).x] + (3/5)^2 - (3/5)^2 -(2/5) = 0 [x-(3/5)]^2 = (2/5)+(9/25) [x-(3/5)]^2 = (19/25) x-(3/5) = √19/5 x = (3/5) ± √19/5 Roots of the equation are x = [3+(√19)]/5 & x = [3-(√19)]/5 |
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| 2911. |
1) 54(m) 1051M 1233. Among two supplementary angles, the measure of the larger angle is 36 more than themeasure of the smaller. Find their measures. |
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| 2912. |
123+123 |
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Answer» by basic addition 123 + 123 = 246- By the process of basic addition 123 + 123 = 246. |
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| 2913. |
123+123-90 |
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Answer» 123+123-90 = 246-90 = 156 |
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| 2914. |
123+123-340 |
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Answer» -94 is the answer of ur question |
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| 2915. |
\left. \begin{array} { l } { \text { Simplify } ( 8 x ^ { 4 } - 5 x - 5 x ^ { 2 } + 8 ) - 7 x ^ { 3 } - 7 x ^ { 2 } } \\ { + ( 2 x ^ { 3 } - 7 x ^ { 2 } + 2 x - 9 ) } \end{array} \right. |
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| 2916. |
DafferantialeS0 |
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Answer» Apply u/V. rulehenced/dx e^x/sinxsinx(d/dxe^x)-e^xd/dxsinx/sin^2x=e^xsinx-e^xcosx/sin^2x |
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| 2917. |
((3*x^2 - x - 2)/(x^2 - 7*x %2B 12))/(((3*x^2 - 7*x - 6)/(x^2 - 4))) |
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| 2918. |
\frac { 3 x ^ { 2 } - x - 2 } { x ^ { 2 } - 7 x + 12 } \div \frac { 3 x ^ { 2 } - 7 x - 6 } { x ^ { 2 } - 4 } |
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| 2919. |
x ^ { 2 } + 7 x + \sqrt { x ^ { 2 } + 7 x + 9 } = 3 |
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| 2920. |
b^(-4)/(2/5)^(-8)=(2/5)^(2*x) |
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Answer» (2/5)^-4 ÷ (2/5)^-8 = (2/5)^2x (2/5)^[-4 - (-8)] = (2/5)^2x (2/5)^4 = (2/5)^2x 2x = 4 x = 4/2 x = 2 (2/5)^-4-(-8)=(2/5)^2x=(2/5)^-4+8=(2/5)^2x=(2/5)^4=(2/5)^2x=4 = 2x(as bases are same so the powers too will be same(=X =4/2= X =2 |
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| 2921. |
\sqrt { 3 x ^ { 2 } - 7 x - 30 } + \sqrt { 2 x ^ { 2 } - 7 x - 5 } = x + 5 , |
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| 2922. |
Extda - Cd tx) |
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Answer» x+d-d-xx-x+d-d0+00 the correct answer is 0 ( x + d ) - ( d + x )= x + d - d - x = x - x + d - d= 0 + 0.please like my answer and accept as best☺☺😊😊😀😀😁😁 x+d-d-x=x-x+d-d=0+0=0 ×+d-f-xx-x+d-d0+00 Answer your question (x+d)-(d-x) Final Result : 2x (x+d) - (d-x) =x+d-d-x=0 ( x + d ) - ( d + x ) = x + d - d - x= x - x + d - d= 0 + 0= 0... PLEASE LIKE AND ACCEPT AS THE BEST... |
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| 2923. |
62e2-3-TX |
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| 2924. |
7A solid is in the form of a right circular cylinder, with a hemisphere at one end and a coneat the other end. The radius of the common base is 3.5 cm and the heights of the cylindricaand conical portions are 10 cm and 6 cm, respectively. Find the total eurface anee of siesolid.1.(Use Ď =3.14) |
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| 2925. |
(5) 63x% + Tx = B |
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Answer» 6√3x^2+9x-2x-√33√3(2x+√3)-1(2x+√3)x=- √3/2 and 1/√3*3 |
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| 2926. |
рдирд╣ 20-Tx+3=0" |
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Answer» 2x² - 7x + 3 = 0 2x² - 6x - x + 3 = 0 2x ( x - 3) - ( x - 3) = 0 (2x - 1) ( x - 3 ) = 0 So, x = 1/2 and 3 |
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| 2927. |
solid is in the form of a right circular cylinder with a hemisphere at one end and athe other end. The radius of the common base is 8 cm. and the heights of the cylindricaland conical portions are 10 cm and 6 cm respectivly. Find the total surface area of thesolid. [use Tt 3.141cone a |
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| 2928. |
18. A water pipe has base radius 6 cm and height 7 cm.What volume of water does it take to fill the pipe? |
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| 2929. |
Tx-15y= 2x+2v = 3 |
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Answer» Multiplying second equation by 7,7x-15y=27x+14y=21 Subtracting second equation from first,-29y=-19y=19/29 x=3-2(19/29)x=(87-38)/29x=49/29 |
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| 2930. |
(1/2)^(-2) %2B (1/3)^(-2) %2B (1/4)^(-2) |
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Answer» 2^2 + 3^2 + 4^24+9+1629 taking inverse ,4+9+16=29 =2^2+3^2+4^2=4+6+8=10+8=18 29 is the correct answer for this question 29 is the right answer..... 29 is the right answer 29 is the correct answer |
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| 2931. |
5^(-2)*((1/3)^(-2) %2B (1/4)^(-2)) |
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Answer» answer of this question is 1 [(3)2+(4)2]x(1/5)29+16 x 1/2525 x 1/251x1 =1 [(3)2+(4)2]×(1/5)29+16×1/2525×1/251×1=1 |
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| 2932. |
2^(2*x %2B 1)=4^(2*x - 1) |
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| 2933. |
2B v (âTx 2y)+9x2y |
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Answer» 5x²y + (-7x²y) + 9x²y - times + is - 5x²y - 7x²y + 9x²y 14x²y - 7x²y 7x²y thanks sir |
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| 2934. |
(-1/5 %2B 1/4)^2/((5/8)) |
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Answer» your answer is 1/250 I am right so please give me a heart also comment |
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| 2935. |
(1/4)^(-2) %2B (1/2)^(-3) %2B (1/3)^(-2) |
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| 2936. |
π×7×7×10×3Tx 3.5x 3.5x10 |
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Answer» N=(π x 7 x 7 x 10 x 3)/(π x 3.5 x 3.5 x 10)N=2 x 2 x 3N=4 x 3N=12 thnks |
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| 2937. |
3x-2y=2Tx + 3 e 43 |
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Answer» Value of x= 4 and y= 5 HIT THE LIKE BUTTON ✔️👍✔️ |
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| 2938. |
7. Two cones of the same base radius 8 cm and height 15 cm are joined together along theirNCERT Exemplar)bases. Find the surface area of the shape so formed. |
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| 2939. |
21. From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of heightsbase radius 6 cm is hollowed out. Find the volume of the remaining solid correct to I place of detrianglecaOr |
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| 2940. |
2^(-3) %2B (2^(n %2B 4) - 2.2^n)/2.2^(n %2B 3) |
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Answer» 0 kaise aaya |
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| 2941. |
(-2*2^n %2B 2^(n %2B 4))/((2*2^(n %2B 3))) |
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Answer» ans 2^n+4 - 2(2^n)/ 2(2^n+3) = 2^n*2^4 - 2(2^n)/ 2(2^n*2^3) = 2^n(2^4 - 2) 2^n(2^4) = (16 - 2)/16 = 14/16 = 7/8 |
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| 2942. |
(2^(n %2B 1)*4^(2*n))/((2^(n - 3)*4^(2*n %2B 1))) |
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| 2943. |
(2) Parallel lines do not intersect'. Write the converse of the statement. Also writewhether the converse is true or not. |
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Answer» Intersecting lines are never parallel to each other. Yes it is true. |
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| 2944. |
थक 2:५3:यदि x = \/—7+ 43 तो ऋन--कक |
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Answer» hit like if you find it useful 4 |
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| 2945. |
Solve for θ :2sin'e+ sin2θ-23-2cosO _ 4sin0-cos2θ + sin28-0.dyOR |
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Answer» 2sin^2x+ (2sinx.cosx)^2=2 2sin^2x+4sin^2x.cos^2x=2 sin^2x+2sin^2x.cos^2x=1 2sin^2x.cos^2x= 1-sin^2x 2sin^2x.cos^2x= cos^2x 2sin^2x=1 sin^2x= 1/2 sinx= 1/√2 => x= pi/4 or 45 degrees |
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| 2946. |
IN CLASS EXERCISEQi,lfaand β are the roots ofthe equation x21-1=0,then find the equation witha2 and B2 as its roots |
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| 2947. |
Check whether the equation 5x2-6x-2 = 0 has real roots and if it has, findthe method of completing the square. Also verify that roots obtained satisfy the giverthem byĐĄĐĄequation |
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| 2948. |
(6) The equation 5x2 -26x+ p 0 has reciprocal roots. Findp.quation 5x2 -26x+ p-0 has reciprocal roots. Find p. |
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Answer» = Let the roots be a and 1/a.= Sum of roots = -b/a= a + 1/a = 26 / 5 = 5 a^2 + 5 = 26 a = 5 a^2 - 26 a + 5 = 0 Solve for a and substitute in place of x to get the value of p./ here one root is a then another root is 1\a so there product is 1=p/5 so p =5 |
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| 2949. |
317) Find the values of p and q for which andχ#-2are the roots ofthe equation pr.gx-620.32 |
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Answer» f(x)=px^2 +qx -6=0f(3/4)=p(3/4)^2 +q(3/4) -6=0=>9p/16 + 3q/4 -6=0=>9p +12q=96.....(1)f(-2)=p(-2)^2+q(-2)-6=04p-2q=6........(2)(2)×6 => 24p -12q =36....(3)(1)+(3)33p =132p=4then q=5 |
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| 2950. |
The roots ofthe equation a(b-c)x-+b(c-a)x+c(a-b)=0are |
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Answer» A,b,c, are 0 each.a,b,c (real number) thanks nehasharma |
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