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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.
| 35451. |
Solutions of the exercise 13.5 question 1 to 4 |
| Answer» Get NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 35452. |
2+5+5= 250 |
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Answer» + ko 4 banado 245 +5=250 How??? |
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| 35453. |
If the HCF of 210 and 55 is expressible in the form 210X+55Y ,find X and Y using EDL |
| Answer» 1 | |
| 35454. |
If sinA+cosB=Under root2 and b=under root 2(sinA+cosA)/sinAcosA, then b is equal to? |
| Answer» | |
| 35455. |
Date sheet 2019 |
| Answer» WWW.CBSE.NIC.COM??? | |
| 35456. |
Lab manual math activity |
| Answer» | |
| 35457. |
Find distance (2,3) (4,1) |
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Answer» 2√2 2√2 |
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| 35458. |
Father of mathematics |
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Answer» Archimedes is the father of mathematics Archimedes Srinivas ramanujan Archimedes.... |
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| 35459. |
Prove that : (cos A + cosecA -1)/(cot A - cosec A + 1 ) = (1+ cos A) / sin A |
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| 35460. |
If sin+ cos=2 to cos - sin= ? |
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Answer» 2(1 - sin theeta) 0 |
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| 35461. |
Write down the distance formula ? |
| Answer» PQ=√(x2-x1) 2+(y2- y1) 2 | |
| 35462. |
Cheak if 231and 396 are coprime |
| Answer» No they are not coprime | |
| 35463. |
If 7 times the 7th term of an Ap is equal to 11times its 11th term, then find its 18th term. |
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Answer» 18th term can also be 0 Thus question is incomplete |
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| 35464. |
Board ki date sheet kab aege |
| Answer» Last December & Start January | |
| 35465. |
For what value of k |
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Answer» oh!god it is an incomplete question. Constant |
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| 35466. |
How to learn trigonometric ratio |
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Answer» learn values of sin then reverse it you get values of cos. For tan use sin÷cos and reverse it to get cot. For cosec use 1/sin and reverse it to get sec. U can learn trigonometric ratio by opening of math book ??? U can learn trigonometric ratio by book |
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| 35467. |
what is the excreation? |
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Answer» The process of removal of toxic substances from an organisms body is called excretion it is the process of removal of metabolic wastes from body |
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| 35468. |
trignomentary questions |
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Answer» trignomentary it is the type of ratios using to solve right triangle problems Please refer to ncert solutions site they are the best ✌ |
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| 35469. |
Tips to prepare for anthe exam |
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Answer» Correct ??? Keep faith in yourself ???? everything will be OK |
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| 35470. |
If the product of zeros of the polynomial ax sq.-6x-6 is 4 find the value of a |
| Answer» a = -3/2 | |
| 35471. |
Show that one and only one out of n,n+3,n+6 is divisible by 4, Where n is any positive integer |
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| 35472. |
Prove that1/secA- tanA-1/cosA=1/cosA-1/secA+tanA |
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Answer» L.H.S=R.H.S(hence proved) Hence proved |
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| 35473. |
Cot theta + cosec theta + 1 by cot theta minus Cos sec theta + 1 |
| Answer» 1 + cos theta by sin theta | |
| 35474. |
Maths wat type of questions will be asked |
| Answer» You are giving borld exam | |
| 35475. |
Tangents make angle |
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Answer» 90degree from radius to the point of contact 90 degree |
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| 35476. |
1/2a+b+2x=1/2a+1/b+1/2x find the value of x |
| Answer» {tex}\\frac{1}{2a + b + 2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0+\xa0{tex}\\frac{1}{2x}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2a + b + 2x}{/tex}\xa0-\xa0{tex}\\frac{1}{2x}{/tex}\xa0=\xa0{tex}\\frac{1}{2a}{/tex}\xa0+\xa0{tex}\\frac{1}{b}{/tex}\xa0{tex}\\Rightarrow{/tex}{tex}\\frac { 2 x - 2 a - b - 2 x } { ( 2 a + b + 2 x ) ( 2 x ) }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2a \\times b}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { - ( 2 a + b ) } { ( 2 a + b + 2 x ) 2 x }{/tex}\xa0=\xa0{tex}\\frac{b + 2a}{2ab}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac { - 1 } { 4 a x + 2 b x + 4 x ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac{1}{2ab}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax = -2ab{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}4x^2 + 2bx + 4ax + 2ab = 0{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0{tex}2x(2x + b) + 2a(2x + b) = 0{/tex}{tex}\\Rightarrow{/tex}\xa0(2x + b)(2x + 2a) = 0{tex}\\Rightarrow{/tex}\xa0x = -{tex}\\frac{b}{2}{/tex} or x = -a | |
| 35477. |
Evaluate:cot45-cos60-sin60/tan45+cos30-sin30 |
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| 35478. |
Ncrt math chpter 12 example 3 solution |
| Answer» Area of segment = Area of sector - Area of triangle Area of triangle = 1/2 r^2 sin theta | |
| 35479. |
Optional ex. Will come in paper or not |
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Answer» Yes , optional questions are important ?\u200d??\u200d? Optional questions never comes in mathematics Anything can come just do your best preparartion ...✌✌ No |
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| 35480. |
How much does the handycam cost if discount 20% |
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Answer» It will cost Rs 19193.6 after 20% discount 23992 |
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| 35481. |
ax+by =a2 , bx +ay=b2 . solve by cross multiplication method |
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Answer» We can write the given system of equations asax + by - a2 = 0........(1)bx + ay - b2 = 0........(2)From equation (1) & (2), we havea1 = a, b1 = b, c1 = -a2a2 = b, b2 = a, and c2 = -b2Therefore, by cross-multiplication, we get{tex}\\Rightarrow \\frac{x}{{b \\times ( - {b^2}) - ( - {a^2}) \\times a}}{/tex}{tex} = \\frac{{ - y}}{{a \\times ( - {b^2}) - ( - {a^2}) \\times b}}{/tex}{tex}= \\frac{1}{{a \\times a - b \\times b}}{/tex}{tex}\\Rightarrow \\frac{x}{{ - {b^3} + {a^3}}} = \\frac{{ - y}}{{ - a{b^2} + {a^2}b}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}Now,{tex}\\frac{x}{{ - {b^3} + {a^3}}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{{a^3} - {b^3}}}{{{a^2} - {b^2}}}{/tex}{tex} = \\frac{{(a - b)\\left( {{a^2} + ab + {b^2}} \\right)}}{{(a - b)(a + b)}}{/tex}{tex} = \\frac{{{a^2} + ab + {b^2}}}{{a + b}}{/tex}also{tex}\\frac{{ - y}}{{ - a{b^2} + {a^2}b}} = \\frac{1}{{{a^2} - {b^2}}}{/tex}{tex}\\Rightarrow - y = \\frac{{{a^2}b - a{b^2}}}{{{a^2} - {b^2}}}{/tex}{tex}\\Rightarrow y = \\frac{{a{b^2} - {a^2}b}}{{{a^2} - {b^2}}}{/tex}{tex} = \\frac{{ab(b - a)}}{{(a - b)(a + b)}}{/tex}{tex} = \\frac{{ - ab(a - b)}}{{(a - b)(a + b)}}{/tex}{tex} = \\frac{{ - ab}}{{a + b}}{/tex}Therefore, {tex}x = \\frac{{{a^2} + ab + {b^2}}}{{a + b}},\\;y = \\frac{{ - ab}}{{a + b}}{/tex}\xa0is the solution of the given system of the equations. bilkul sahi answer hai |
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| 35482. |
the 6th term of AP is 0 prove that 31st term is 5 times the 11th term |
| Answer» a6=0 to prove=a31=5(a11)proof=a31=a +30da11=a +10da+30d=5(a+10d)=a+30d=5a+50d4a+20d=0 i.e a+5d=0we havea6=0a+5d=0 | |
| 35483. |
2_2_2_2_2=20 you do (+,-,÷,×) |
| Answer» 2+2x2+2x2=20 | |
| 35484. |
Find two consecutive number whose squares have the dum 85 |
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Answer» (x+2x)square =85×square+4xsquare=85×square+4xsquare-85=0Now solve it by splitting method and you will get 2 value of x then put these values in the given consecitive no.and you will get the two Consecutive no.s .Thanks 2square+9square |
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| 35485. |
Class 10 sample paper |
| Answer» Check Sample Papers here : https://mycbseguide.com/cbse-sample-papers.html | |
| 35486. |
Find the product of cos 30°.cos 45°.cos 60° |
| Answer» cos30 = 1/2cos 45 = 1/\xa0√2cos 60 = √3/2 | |
| 35487. |
Show that one number n, n+1and n+4 is divisible by 3 |
| Answer» Take n=3q Then find | |
| 35488. |
What is a chord |
| Answer» A line join to poin of a circle exist inside the circle | |
| 35489. |
Prove that sin0 - cos0 =1 |
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Answer» sin 0 = 0cos 0 = 1sin 0 - cos 0 = 0 - (1) = -1 sin0=0cos0=1So your question is wrong Sin0- COS0=1 |
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| 35490. |
Find the roots of 1/x - 1/x - 2= 3 |
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Answer» Kon nhi bta rha h ji Nhi bta rhe jao bhagho |
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| 35491. |
Find the h.c.f.of the terms and factories 5y-15y\'2 |
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| 35492. |
2√5/2 is irrational prove? |
| Answer» | |
| 35493. |
1/2+1/2+3/4 |
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Answer» 1/2+1/2=1and 1+3/4=1.75 so, the answer was 1.75 1.75 1.7 |
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| 35494. |
Formula of median? |
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Answer» L+(n/2-c.f)×h Median=lower limit +[n÷2-cumulative frequency whole ÷by frequency ]×h |
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| 35495. |
Check whether fhe following are quadratic equation( x+1)2=2(x-3) |
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Answer» No, it is not a quadratic equation No it is not a quadratic equation X |
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| 35496. |
Can (x-7) be the remainder on division of a polynomial p(x) by 7x+2?justify your answer |
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| 35497. |
3t-2/3 + 2t+3/2 = t+7/8 |
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| 35498. |
If the areas of two similar triangle are equal , prove that the 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} | |
| 35499. |
Prove that sin theta /1+cos +1+cos theta/sin theta =2 cosec theta |
| Answer» Pls take lcm and pls factorise and cancel out cos theta u will get the answer | |
| 35500. |
SinA cosecA =? |
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Answer» 1 1 SinA. CosecA = sinA. 1/sinA = 1 (sinA and sinA gets cancelled) 1 SinA cosec A = sin A × 1/sinA = 1 |
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