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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.
| 10901. |
PS is The bisector of |
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Answer» . Where\'s your question???? ????? |
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| 10902. |
Prove that √5 is irrational |
| Answer» Let take √5 as a rational numberIf a and b are two co-prime number and b is not equal to 0.We can write √5 = a/bMultiply by b both side we getb√5 = aTo remove root, Squaring on both sides, we get5b2 = a2 ……………(1)Therefore, 5 divides a2 and according to a theorem of rational number, for any prime number p which is divided \'a2\' then it will divide \'a\' also.That means 5 will divide \'a\'. So we can writea = 5cand putting the value of a in equation (1) we get5b2 = (5c)25b2 = 25c2Divide by 25 we getb2/5 = c2again using the same theorem we get that b will divide by 5and we have already get that a is divided by 5but a and b are co-prime number. so it is contradicting.Hence √5 is an irrational number | |
| 10903. |
What is Probability |
| Answer» Many events cannot be predicted with total certainty. the best way is to say How likely some\xa0event is to happen using the idea of probabilitye.g tossing a coin , there are two posible outcomeswe can get either head or tailprobability of getting head is 1/2probability of getting tail is 1/2 | |
| 10904. |
What is teta |
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Answer» For not four Sorry data not teta data means for molecules of an element |
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| 10905. |
Why maths are not in hindi |
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| 10906. |
what is an formula for mode |
| Answer» Mode= l+f-f1/2f-f1-f2*h | |
| 10907. |
X^2 +3x+4=0 |
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| 10908. |
using prime factorization find hcf and lcm of 72,126,168 |
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| 10909. |
2+2+2+2+2+= |
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Answer» 10 10 |
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| 10910. |
Square root |
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| 10911. |
T×T-15 |
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| 10912. |
5/5 =1 |
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| 10913. |
Thi |
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| 10914. |
If one zero of polynomial ( a2+q)x2+βx+6a is reciprocal of other find value of a. |
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| 10915. |
Chapter 8 all identity write and give me |
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| 10916. |
1+ tan2A = |
| Answer» Sec2A | |
| 10917. |
How many terms of the a.p. 9,17,25,....must be taken so that so their sum is 636 |
| Answer» a=9,d=8,n=?,sn=636So sn=n/2(2a+n-1)d 636=n/2(18+n-1)8 313=n(144+8n-8) 313=n(136+8n) 313=136n+8n² 8n²+136n-313. | |
| 10918. |
Triangle ABC is right angled at a A then the value of tanB.tanC is |
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Answer» tan B = AC/AB tan C = AB/ACso tan B * tan C = AC/AB*AB/AC =1 1 1 |
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| 10919. |
If sina + 2cosa =1 then prove 2sina - cosa =2 . |
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| 10920. |
The graph of y=P (X) IS GIVEN. Find the number of zeros of PX |
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| 10921. |
Show that 3/125 is terminating rational number |
| Answer» Prime factorizatio of 125=5×5×5Therefore it is terminating rational no. | |
| 10922. |
Express the HCF of numbers 72 and 124 as a linear combination of 72 and 124 |
| Answer» 124×7-72×13 | |
| 10923. |
cosA+sinA= equal to root 2 cos theta |
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| 10924. |
what is frustum |
| Answer» Frustum is\xa0the portion of a cone or pyramid which remains after its upper part has been cut off by a plane parallel to its base, or which is intercepted between two such planes. | |
| 10925. |
find HCF of 3/5,7/5 &1/2 |
| Answer» Hcf of 3,7,1=1 now LCM of 5,5,2=10. Hcf of 3/5,7/5&1/2=1/10 | |
| 10926. |
1÷3x+Y + 1÷3x-y=3÷4 |
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| 10927. |
If d is the hcf of 210 and 455, find x and y satisfying d = 210x + 55y. Plz answer fastly... |
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| 10928. |
sin of trigo |
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| 10929. |
if sin(A+B)=1/2 and cos(A-B)=1/2 find A and B |
| Answer» Sin 30=1/2 so A+B=30---------eq1Cos60=1/2 so A-B=60---------eq2Now solve eq 1and 2A+B=30A-B=602A =90A=45B=-15Ok | |
| 10930. |
In an acute angled triangle ABC, if tan(A+B-C) =1 and ,sec(B+C-A) =2,find the value of A, B andC |
| Answer» According to the question, tan (A+B-C) = 1\xa0{tex}\\Rightarrow{/tex}\xa0tan (A+B-C) =\xa0{tex}\\tan 45 ^ { \\circ }{/tex}{tex}\\Rightarrow{/tex}\xa0A + B - C = 45° ........(1)Also given, sec (B+C-A) = 2{tex}\\Rightarrow{/tex}\xa0sec (B + C - A ) = sec 60°\xa0{tex}\\therefore{/tex}{tex}B + C - A = 60 ^ { \\circ }{/tex}...........(2)Adding equation (1) & (2);{tex}( A + B - C ) + ( B + C - A ) = 45 ^ { \\circ } + 60 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad 2 B = 105 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad B = 52 \\frac { 1 ^ { \\circ } } { 2 }{/tex}Putting\xa0{tex}B = 52 \\frac { 1 ^ { \\circ } } { 2 }{/tex}\xa0in equation (2); we get :-{tex}52 \\frac { 1 ^ { \\circ } } { 2 } + C - A = 60 ^ { \\circ }{/tex}{tex}\\Rightarrow C - A = 7 \\frac { 1 ^ { \\circ} } { 2 }{/tex}......(3)Also, in\xa0{tex}\\Delta A B C{/tex}, we have\xa0{tex}A + B + C = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad A + 52 \\frac { 1 ^ { \\circ } } { 2 } + C = 180 ^ { \\circ }{/tex}\xa0{tex}\\left[ \\because B = 52 \\frac { 1 ^ { \\circ } } { 2 } \\right]{/tex}{tex}\\Rightarrow \\quad C + A = 127 \\frac { 1 ^ { \\circ } } { 2 }{/tex}......(4)Adding and subtracting (3) and (4), we get\xa02C = 135o and 2A = 120o{tex}\\Rightarrow C = 67 \\frac { 1 } { 2 } ^ { \\circ } {/tex}\xa0and A = 60oHence, we get the values of A = 60o,{tex}B = 52 \\frac { 1 ^ { \\circ } } { 2 }{/tex}and\xa0{tex}C = 67 \\frac { 1 ^ { \\circ } } { 2 }{/tex}.\xa0 | |
| 10931. |
For the value of 5 |
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| 10932. |
Solvea÷bx-1+b÷ax-1=a+b |
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| 10933. |
A÷bx-1+b÷ax-1=a+b |
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| 10934. |
Solve a÷bx-1+b÷ax-1 |
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| 10935. |
Sin A |
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Answer» It can be Tan A × Cos A oropp/hypo it can be Sin A × Cos A oropp/hypo We can also write sin A =0 Sin A =1/cosecA |
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| 10936. |
Draw a line segment of length 7.6cm and divide in the ratio5:8.Measure the two parts |
| Answer» Given: A line segment of length 7.6 cm.Required: To divide it in the ratio 5 : 8 and to measure the two parts.Steps of construction :\tFrom any ray AX, making an acute angle with AB.\tLocate 13 (= 5 + 8) points A1,\xa0A2, A3,..... and A13\xa0on AX such that\tAA1\xa0= A1A2\xa0= A2A3\xa0= A3A4\xa0=\xa0A4A5\xa0= A5A6\xa0= A6A7\xa0= A7A8\xa0= A8A9\xa0= A9A10\xa0= A10A11\xa0= A11A12\xa0= A12A13\tJoin BA13\tThrough the point A5, draw a line parallel to A13B intersecting AB at the point C.\tThen, AC : CB = 5 : 8\tOn measurement, AC = 3.1 cm, CB = 4.5 cm.Justification :{tex}\\because{/tex}\xa0A5C || A13B [ By Construction]{tex}\\because{/tex}\xa0{tex}\\frac { A A _ { 5 } } { A _ { 5 } A _ { 13 } } = \\frac { A C } { C B }{/tex}\xa0[By the Basic proportionality theorem]But,\xa0{tex}\\frac { A A _ { 5 } } { A _ { 5 } A _ { 13 } } = \\frac { 5 } { 8 }{/tex}\xa0[ By Construction[Therefore,\xa0{tex}\\frac { A C } { C B } = \\frac { 5 } { 8 }{/tex}This shows that C divides AB in the ratio 5 : 8. | |
| 10937. |
How to now acid and base |
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Answer» You can use litmus paper, ph scale or any universal indicator to find whether the given compound or element is acid or base. When acid is added in water then hydronium ion is released and when Base is added in water then hydroxide ion is produced. You can use Indicators to Determine acids and bases Acids are sour in taste and bases are bitter in taste and also by using indicators Acid is sour in taste and Base is better in taste |
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| 10938. |
Evaluate (-1)3*(-2)3*(-3)3*(-4)3*2(5)3 |
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| 10939. |
If x^2 + y^2 + 16z^2 = 48, xy + 4yz + 2zx = 24. Find x^2 + y^2 + z^2. |
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| 10940. |
Show that any positive integer is of the form 3q or 3q+1 or 3q+2, for some integer q |
| Answer» It proven by euclids lemma use a= bq + r put value b 3 q = b and r = 0,1,2 | |
| 10941. |
Find the root by factorization 2x²—x+1/8=0 |
| Answer» We have, {tex}2x^2-x+{1\\over8}=0{/tex}{tex}\\implies 2x^2-{1\\over2}x-{1\\over2}x+{1\\over8}=0{/tex}{tex}\\implies x(2x-{1\\over2}) - {1\\over4}(2x-{1\\over2})=0{/tex}{tex}\\implies (2x-{1\\over2}) (x-{1\\over4})=0{/tex}{tex}Either \\,(2x-{1\\over2})\\, =0\\,or\\, (x-{1\\over4})=0{/tex}{tex}\\implies x = {1\\over4},\\,{1\\over4}{/tex}So, this root is repeated root.{tex}\\therefore {/tex} both the roots are {tex}{1\\over4}{/tex}. | |
| 10942. |
B×x÷a + a×y÷b =a^2 +b^2 and x+y=2 ab solve the equation by substitution method |
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| 10943. |
What is the meaning of periods in sylllabus |
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| 10944. |
Find the value of p such that the quadritic equation (p_12)x2_2 (p_12)x+220 has equal root |
| Answer» 245 | |
| 10945. |
Find the HCF of 426 and 576 and express it in the form of 52x +117y . |
| Answer» 75 | |
| 10946. |
HD sdhngdfgh |
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| 10947. |
1-tan^4A/1+tan^4A=cosA+SinA/CosA-SinA....prove |
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| 10948. |
Thmmmb |
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| 10949. |
If (1-sinø) (1+sinø) =1/4 , find the value of tan ø |
| Answer» {tex}(1-sin\\theta)(1+sin\\theta)+{1\\over 4}\\\\=> 1-sin^2\\theta ={1\\over 4}\\\\=> cos^2\\theta={1\\over 4}\\\\=> cos\\theta ={1\\over 2}\\\\=> cos \\theta= cos 60^°{/tex}{tex}=> \\theta =60^°\\\\tan 60^°= \\sqrt 3{/tex} | |
| 10950. |
mam=nan and show am+n=0 |
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