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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.
| 27851. |
In board exam justification of construction is important or not |
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Answer» Depends on question Yes |
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| 27852. |
Coordinates (-3,-14) and (a,-5) is equal distance 9 unit find the value of a |
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| 27853. |
PROVE THAT (sin^8A + cos^8A) = (sin^2A - cos^2A)(1 - 2sin^2Acos^2A) |
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Answer» Question:\xa0PROVE THAT (sin^8A + cos^8A) = (sin^2A - cos^2A)(1 - 2sin^2Acos^2A)Answer: Ya baski nahi h Ye question cbse 2013 ka h |
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| 27854. |
Find the area of a quadrant of a circle whose circumference is 44 cm. |
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Answer» circumference of the circle= 44cm2πr=442×22/7×r=4444/7×r=44r=44×7/44r=7cmar. of quadrant= 1/4πr² =1/4×22/7×7² =77/2 =38.5(hope this is correct) First use circumference then area of quadrant which is 1/4 pie (r)2 then solve easily |
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| 27855. |
(a-b)x + (a + b)y= a² - 2ab-b²(a+b)(x+y)=a²+b² |
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| 27856. |
If the area of two similar trianglesbare equal , prove that they are congruent |
| Answer» Given:{tex}∆ABC\\sim∆PQR{/tex}and\xa0{tex}ar∆ABC=ar∆PQR{/tex}To prove:\xa0{tex}∆ABC\\cong∆PQR{/tex}Proof:\xa0{tex}∆ABC\\sim∆PQR{/tex}\xa0Also\xa0{tex}\\operatorname { ar } ( \\Delta A B C ) = \\operatorname { ar } ( \\Delta P Q R ){/tex}\xa0(given)or,\xa0{tex}\\frac { \\operatorname { ar } ( \\Delta A B C ) } { \\operatorname { ar } ( \\Delta P Q R ) } = 1{/tex}Or\xa0{tex}\\frac{AB^2}{PQ^2}=\\frac{BC^2}{QR^2}=\\frac{CA^2}{RP^2}=1{/tex}Or\xa0{tex}\\frac{AB}{PQ}=\\frac{BC}{QR}=\\frac{CA}{RP}=1{/tex}Hence we get thatAB=PQ,BC=QR and\xa0CA=RPHence\xa0{tex}∆ABC\\cong ∆PQR{/tex} | |
| 27857. |
2+4-6(2-7+7)×8×2+8-0 |
| Answer» -178 | |
| 27858. |
Factories 3k²-16k+14 |
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| 27859. |
Square of 45 |
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Answer» 2025 2025 |
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| 27860. |
6q+1 or 6q+3 or 6q+5 |
| Answer» Its very lengthy to type.....use euclid\'s division lemma...a=bq+r where b=6 and values of r = 1,2,3,4,5 then put the values of r like a=6q(eqn 1) , a=6q+1(eqn2)....a=6q+5(eqn6) then from 2 4 and 6 we can say that any +ve odd integer is in the form of 6q+1, 6q+3, 6q+5 | |
| 27861. |
Math ke kuch cbse board ke liye important question bata do |
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| 27862. |
How can we solve trigonometric identities easily? |
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Answer» Evaluate the identities Trigonometry bhut hi simple ch h aap bs free ho kr uske que solve krna easily ho jaenge But all the questions have different concepts and type of solving so it is difficult to understand which concept will be used where Remember all the identities..... and practice more questions ..... |
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| 27863. |
Important 4mark question |
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| 27864. |
Ex 6.6 |
| Answer» Yes | |
| 27865. |
if 7cosecA-3cotA=7,prove that 7cotA-3cosecA=3 |
| Answer» Take 7 to lhs and 3cotA to rhs.... take 7 common frm lhs... now multiply cotA on both sides........ then put cot sq.A= 1+cosec sq. A in lhs..... then solve it....yu will get the answer | |
| 27866. |
How to convert centimetre square into litre |
| Answer» 100cm sq. = 1 litre..... | |
| 27867. |
how to solve x^2 +400x + 960000 such large quadratic equations |
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Answer» Splitting the middle term X2. + 400x - 960000X2 + ( 1200x - 800x) + ( 1200 × 800)x ( x + 1200) - 800 ( x + 1200)(x + 1200)(x - 800) First findb2-4ac=400*400-4(960000)=160000-3840000= negativeSo the solution of this equation is not possible Take it 10^n for that Use quadratic formula or completing the square |
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| 27868. |
Hcf of 612 and 408 |
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Answer» Prime factors of 408 = 23 × 3 × 17 Prime factors of 612 = 22 ×32 × 17Common prime factors: 2 , 3 , 17HCF = 22×31× 171 = 204 612=408×1+204....408=204×2+0........Therefore, HCF of 612 and 408 is 204 |
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| 27869. |
37x+ 43y43x+ 37y |
| Answer» CBSE Sample Paper for Class 11 English (Solved) – Set D | |
| 27870. |
All exercise |
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| 27871. |
Find the greatest number of 5 digits exactly divisible by 12, 15 and 36 |
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Answer» L.C.M of 12‚15‚36 = 180 Largest 5 digit number =99999 99999 = 180 × 555 + 99 Largest 5 digit number divisible by 12‚15‚36= 99999 - 99=99900 99900 is the correct answer ChupTMC In this take LCM of all no. Which is 180.So 180 is the answer. To be sure of answer u can also divide 180 with all no. They divide 180 |
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| 27872. |
Let 3 n0. Are x-1 ,x+2 x-7are in AP find the value of x |
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| 27873. |
Ssum of 2 digit no |
| Answer» Continue ur ques... | |
| 27874. |
Sin 6 theta + cos 6 theta + 3 sin square theta cos a cos square theta equals to 1 |
| Answer» {tex}\\sin ^ { 6 } \\theta + \\cos ^ { 6 } \\theta = 1 - 3 \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta{/tex}{tex}\\text { L.H.S. } = \\sin ^ { 6 } \\theta + \\cos ^ { 6 } \\theta{/tex}{tex}= \\left( \\sin ^ { 2 } \\theta \\right) ^ { 3 } + \\left( \\cos ^ { 2 } \\theta \\right) ^ { 3 }{/tex}{tex}= \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta \\right) \\left( \\sin ^ { 4 } \\theta + \\cos ^ { 4 } \\theta - \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta \\right){/tex}{tex}\\left[ \\because a ^ { 3 } + b ^ { 3 } = ( a + b ) \\left( a ^ { 2 } + b ^ { 2 } - a b \\right) \\right]{/tex}{tex}= 1 \\left( \\sin ^ { 4 } \\theta + \\cos ^ { 4 } \\theta + 2 \\sin ^ { 2 } \\theta \\cdot \\cos ^ { 2 } \\theta - 2 \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta - \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta \\right]{/tex}{tex}\\text { since, } \\sin ^ { 2 } A + \\cos ^ { 2 } A = 1{/tex}{tex}= \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta \\right) ^ { 2 } - 3 \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta{/tex}{tex}= 1 - 3 \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta{/tex}= R.H.S. proved. | |
| 27875. |
P.T -1+ sinA.sinA(90-A)/ cot(90-A) = -sin²A |
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| 27876. |
Pathageras theorem proof |
| Answer» Ncert me dekho | |
| 27877. |
Bpt theirm |
| Answer» see in ncert books page no. 124 | |
| 27878. |
Show that n square -1 is divisible by 8, if n is a positive integer.Explain breifly guys plzz??? |
| Answer» The question is like this:Show that n2\xa0- 1 is divisible by 8, if n is an odd positive integerSoln:Any odd positive integer\xa0n\xa0can be written in form of 4q\xa0+ 1 or 4q\xa0+ 3.\xa0If\xa0n\xa0= 4q\xa0+ 1, then\xa0n2\xa0- 1 = (4q\xa0+ 1)2\xa0- 1 = 16q2\xa0+ 8q\xa0+ 1 - 1 = 8q(2q\xa0+ 1) which is divisible by 8.If\xa0n\xa0= 4q\xa0+ 3,then\xa0n2\xa0- 1 = (4q\xa0+ 3)2\xa0- 1 = 16q2\xa0+ 24q\xa0+ 9 - 1 = 8(2q2\xa0+ 3q\xa0+ 1) which is divisible by 8.\xa0So, n2\xa0- 1 is divisible by 8, if n is an odd positive integer. | |
| 27879. |
Please explain chapter-12 example-4 |
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Answer» Example ka solution dekha bina solve karo nahi to confuse hoga Pahle koi ek segment ka area nikalo usse 2 sa multiple karo aur fir usmain area of square add kar do ans aajaayega |
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| 27880. |
Is it necessary to write both step of construction and justification for 4 marks?????.. |
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Answer» No...justify if needed bt write const Yes Yes |
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| 27881. |
Prove that tangents drawn at the end points of a chord of a circle make equal angles with the chord. |
| Answer» In Triangle APBSince\xa0PA = PB (Tangents from same external point are euqal)\xa0{tex}\\therefore {/tex}\xa0{tex}\\angle P A B = \\angle P B A{/tex}(Opposite angles to\xa0equal sides are equal) | |
| 27882. |
Prove underroot 3 |
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| 27883. |
Divides 27 into 2 part such that sum of their reciprocals is 3/4 |
| Answer» question is wrong | |
| 27884. |
Find the distance between the point A(a+b,a-b) and (a-b,-a-b) |
| Answer» 4b^2 | |
| 27885. |
how many places will expansion terminate23/2\'3 5\'3 |
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Answer» 0.00230 Right answer 0.023 is ri8 answer. |
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| 27886. |
without actually performing the long division state whether the rational number 13/343 |
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| 27887. |
PQ and PR are tangent to the circle with centre O such that angleQPR =50, then find angle OQR |
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Answer» 25° Sorry sorry answer is 25 45 hai i think Don\'t know |
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| 27888. |
Find the value of \'k\' so that equations have equal roots Kx(x-2) +6=0 |
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Answer» D=b2- 4ac=0(-2k)2 -4(k)(6)=0(4k)2-4k(6)=016k2-24k=016k2=24kK=6 Kx (x-2)+6=0=> kx^2-2kx+6=0Now this equation has equal rootsTherefore D=0=>b^2-4ac=0=>(-2k)^2-4×k×6=0=>4k^2-24k=0=>4k (k-6)=0=>k=0 or k=6 |
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| 27889. |
Solve for x and yX/a +y/b = 2ax - by = a2- b2 |
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Answer» x/a + y/b = 2ay + xb = 2ab----------( 1 )ax - by = a² - b²---------( 2 )from-----( 1 ) &-------( 2 )multiply ( 1 ) by "a" ( 2 ) by "b"abx + a²y = 2a²babx - b²y = ba² - b³-------------------------------------------------y(a² + b²) = 2a²b - ba² + b³y(a² + b²) = a²b + b³y(a² + b²) = b(a² + b²)y = b [ put in -----( 1 )]we get,ay + xb = 2aba(b) + xb = 2abxb = 2ab - abx =ab/bx=a is that a2 or a^2 |
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| 27890. |
Find a cubic polynomial whose zeros are 1/2,1 and -3 |
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Answer» 2x^3 + 3x^2 -8x + 3 2x³+3x²-4x+3 |
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| 27891. |
prove that √7 is irrational (plz write step by step) |
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Answer» Let root 7 be a rational numberThen root 7 is equal to p upon qwhere p and q are coprime integers and q is not equals to zero........By squaring both sides....7=p²/q²........q²=p²/7If 7 divides p² 7 also divides p.....Let p=7c......q²=49c²/7.........q²=7c².........q²/7=c²......If 7 q² then 7 also divides q......Hence p and q are not coprime integers.....Our assumption was wrong that √7 is a rational number........Root 7 is not a rational number.....Root 7 is an irrational number what types of Questions have chances to come in board xams from ch 1 You can see in text book page 13...there it is proved that root3 is irrational....u can get help from there |
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| 27892. |
√7 is irrational no |
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Answer» Yes Yaa it is Yes |
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| 27893. |
If the zeros of polynomial are 3+√5 and 3-√5 . Find the polynomial |
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Answer» Xsq.-6x+4 x^2 - 6x + 4 |
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| 27894. |
Find the HCF 135 and 225 |
| Answer» 45 | |
| 27895. |
In RD sharma Pg no 11.24 Question no 11 Please |
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Answer» X=2 See at following l ink I don,t have RD sharma book:https://www.topperlearning.com/study/cbse/class-10/mathematics/text-book-solutions/r-d-sharma-mathematics-x/42/trigonometric-identities/277/b101c2s3e9 |
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| 27896. |
Sin^3√®°°°©°°°==©°°°°°®°°°°°=£¢¢¢¢¢®©©©©℅•••• |
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Answer» Ye kya h.....class 10 ke kisi book me nhi milega aisa kha se laya ye sab What is your question i can\'t understand |
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| 27897. |
What things we have to carry in board examination hall |
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Answer» Only your admit card transparent kit only use blue pen in the exam And other necessary things are written on it First of alll .your admitcard Written in admit card Say fast i also don\'t know |
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| 27898. |
Find the sum of all the three digit numbers which leave the remainder 3 when divided by 5 |
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Answer» 99090 Its ap have 103, 108 , 113,118..............198 yeh ap ayge okay Jyoti you ap is wrong AP=103,113,123........193An=a+(n-1)×d193=103+(n-1)×1090=(n-1)×109=n-110=nS10=10/2(a+l)S10=10/2(103+193)S10=5×296S10=1480 |
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| 27899. |
In an Ap, if the common difference = -4,and the seventh term =is 4 then find the first term |
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Answer» Jyoti aapne last me fault krdi a= 28 A7=a+6d4=a+6(-4)4=a-243+24=a27=a |
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| 27900. |
If cos+sin =√2cos,show that cos-sin=√2sin |
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Answer» - mane minus sign ko common lo - ko common le lo fir equate kro answer aa jayega (CosA+sinA)^2=(_/2 cosA)^2Cos^2A+sin^2A+2cosAsinA=2cos^2A1-sin^2A+1-cos^2A+2cosAsinA=2 (1-sin^2A) Squaring both sides krke kr lo ho jayega |
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