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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.
| 26601. |
write an ap a=5 |
| Answer» Here, a = 5, d = 3, an = 50We know thatan = a + (n – 1)d{tex} \\Rightarrow {/tex}50 = 5 + (n - 1)3{tex} \\Rightarrow {/tex} (n – 1)3 = 50 - 5{tex} \\Rightarrow {/tex}\xa0(n - 1)3 = 45{tex} \\Rightarrow n - 1 = \\frac{{45}}{3}{/tex}{tex} \\Rightarrow {/tex}\xa0n - 1 = 15{tex} \\Rightarrow {/tex}n = 15 + 1{tex} \\Rightarrow {/tex}n = 16Again, we know that{tex}{S_n} = \\frac{n}{2}\\left[ {2a + (n - 1)d} \\right]{/tex}{tex} \\Rightarrow {S_n} = \\frac{{16}}{2}\\left[ {2(5) + (16 - 1)3} \\right]{/tex}{tex} \\Rightarrow {/tex}\xa0Sn = 8[10 +45]{tex} \\Rightarrow {/tex}\xa0Sn = 8(55){tex} \\Rightarrow {/tex}\xa0Sn = 440 | |
| 26602. |
How to understand coordinate geometry |
| Answer» By reading | |
| 26603. |
Will rd Sharma questions came for the exam or ncert question |
| Answer» I had completed both | |
| 26604. |
Factor of x²-root2 |
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| 26605. |
Derivation of section formula |
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| 26606. |
state and prove thales theroem |
| Answer» Long | |
| 26607. |
So that exactly one of the number n, n + 2 or n + 4 is divisible by 3 |
| Answer» Let n =3kthen n + 2 = 3k + 2and n + 4 = 3k + 4Case 1: When n=3k ,n is divisible by 3 ............(1)n + 2 = 3k + 2or, n + 2 is not divisible by 3n + 4 = 3k + 4= 3(k + 1) + 1or, n + 4 is not divisible by 3Case 2:When n=3k+1, n is not divisible by 3\xa0n + 2 = (3k + 1) + 2=3k + 3 = 3(k + 1){tex} \\Rightarrow{/tex}\xa0n+ 2 is clearly divisible by 3..........................(2)n + 4 = (3k + 1) + 4= 3k + 5= 3(k + 1) + 2{tex}\\Rightarrow{/tex}\xa0n + 4 is not divisible by 3Case 3:When n=3k+2,n is not divisible by 3\xa0n + 2 = (3k + 2) + 2= 3k + 4(n + 2) is not divisible by 3x + 4 = 3k + 6 = 3(k + 2){tex}\\Rightarrow{/tex}\xa0n + 4 is divisible by 3........................(3)Hence, from (1),(2) and (3) it is clear that\xa0exactly one of the numbers n, n + 2, n + 4, is divisible by 3. | |
| 26608. |
Factor of x²-2 |
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Answer» x2\xa0-2(x+√2)(x-√2) +_underroot 2 |
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| 26609. |
If cos theta-b sin theta=c,prove that a sin theta+b cos theta=+-√a²+b²+c² |
| Answer» We have, {tex}asin\\theta+bcos\\theta=c{/tex}On squaring both sides, we get{tex}(asin\\theta+bcos\\theta)^2=c^2{/tex}(a sin θ)2\xa0+ (b cos θ)2\xa0+ 2(a sin θ) (b cos θ) = c2⇒ a2\xa0sin2\xa0θ + b2\xa0cos2\xa0θ + 2ab sin θ cos θ = c2⇒ a2(1 – cos2\xa0θ) + b2\xa0(1 – sin2\xa0θ) + 2 ab sin θ cos θ = c2 {tex}[\\because sin^2\\theta+cos^2\\theta=1]{/tex}⇒ a2\xa0– a2\xa0cos2\xa0θ + b2\xa0– b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2⇒ –a2\xa0cos2\xa0θ – b2\xa0sin2\xa0θ + 2ab sin θ cos θ = c2\xa0– a2\xa0– b2 Taking Negative common,⇒ a2\xa0cos2\xa0θ + b2\xa0sin2\xa0θ – 2ab sin θ cos θ = a2\xa0+ b2\xa0– c2⇒ (a cos θ)2\xa0+ (b sin θ)2\xa0– 2(a cos θ) (b sin θ) = a2\xa0+ b2\xa0– c2⇒ {tex}(acos\\theta-bsin\\theta)^2=a^2+b^2-c^2{/tex}⇒{tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\pm \\sqrt { a ^ { 2 } + b ^ { 2 } - c ^ { 2 } }{/tex}\xa0Hence proved, {tex}acos\\theta-bsin\\theta{/tex} =\xa0{tex}\\sqrt{a^2+b^2-c^2}{/tex} | |
| 26610. |
Cnggjk |
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| 26611. |
How to do route method |
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| 26612. |
Is cbse sample paper is enough to score 60+ |
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| 26613. |
Find the length of tangent in a circle with radius 6 cm |
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| 26614. |
Draw a line segment of lengths 7.6cm and divide it in the ratio 5:8 measer the two part |
| 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. | |
| 26615. |
X2+2x+48 |
| Answer» | |
| 26616. |
Express the h.c.f of 48 and 18 as a linear combination |
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Answer» A 48 => 2x2x2x2x318 => 2x3x3 48-2×2×2×2×3 |
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| 26617. |
The median of the following data is 28 .find the valves of x and y .if the total frequency is 50 |
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| 26618. |
6x3 + 5x2–12x+4 |
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| 26619. |
Bye to bol diya kro dear...Samruddhi ? |
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| 26620. |
If a,b,c,l,m are in ap find the value of a-4b+6c-4l+m |
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Answer» And so easy ???? Refer to Rd Sharma |
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| 26621. |
RIMSHA plzzz aa jao |
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| 26622. |
RIMSHA where r u dear????? |
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| 26623. |
RIMSHA come dear????? |
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| 26624. |
RIMSHA???????? |
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| 26625. |
Datesheet fr 10th ??? |
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| 26626. |
(8, 1), (k, -4), (2, -5) |
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| 26627. |
Vxec |
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| 26628. |
X-y/2=3 |
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| 26629. |
Cbse date sheet kab aa gyi or not |
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| 26630. |
Give prime factorisation of 4620 |
| Answer» The prime factorization of 4620 is{tex}4620 = 2 \\times 2 \\times 3 \\times 5 \\times 7 \\times 11{/tex}{tex}\\quad = 2 ^ { 2 } \\times 3 \\times 5 \\times 7 \\times 11{/tex} | |
| 26631. |
If \'alpha\' &\'bita are zeros of the:x2+7x+7 then find1/alpha+1/bita-2alphabita |
| Answer» then alpha + bita will be -7 and alpha × buta will be 7 so i upon alpha + i upon bita will be alpha +bita upon alpha ×bita - 2alpha×bita so it would be - 13 | |
| 26632. |
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289,find the sum of first n terms |
| Answer» Let a be the first term and d be the common difference of given AP.Then, we haveS7=49{tex}\\Rightarrow \\frac { 7 } { 2 } [ 2 a + 6 d ] = 49{/tex}{tex}\\Rightarrow \\frac { 7 \\times 2 } { 2 } [ a + 3 d ] = 49{/tex}{tex}\\Rightarrow{/tex}a+3d=7...(i)Also, S17=289{tex}\\Rightarrow \\frac { 17 } { 2 } [ 2 a + 16 d ] = 289{/tex}{tex}\\Rightarrow \\frac { 17 \\times 2 } { 2 } [ a + 8 d ] = 289{/tex}{tex}\\Rightarrow{/tex}a+8d=17....(ii)Subtracting (i) from (ii), we get,5d=10{tex}\\Rightarrow{/tex}d=2{tex}\\Rightarrow{/tex}a=7-3(2)=7-6=1{tex}\\therefore{/tex}{tex}S _ { n } = \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}{tex}= \\frac { n } { 2 } [ 2 ( 1 ) + ( n - 1 ) 2 ]{/tex}{tex}= \\frac { n } { 2 } [ 2 + 2 n - 2 ]{/tex}={tex}\\frac{n}{2}{/tex}{tex}\\times{/tex}2n=n{tex}^2{/tex} | |
| 26633. |
Solve for X and y x+y=a+byax-by =a sq-b sq |
| Answer» x + y = a - b ............. (i)and ax - by = a2 + b2 ............... (ii)Multiply (i) by b and subtract by (ii)x = aPut x = 1 in (i)x + y = a - ba + y = a - by = -b | |
| 26634. |
Solve for X and y |
| Answer» | |
| 26635. |
Find the value of k for which one root of the quadratic equation kx^2-14x+8=0 is six times the other |
| Answer» Let one root = {tex}\\alpha{/tex}Other root = 6{tex}\\alpha{/tex}{tex}\\therefore{/tex}Sum of roots =\xa0{tex}\\alpha{/tex}\xa0+ 6{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{14}{k}{/tex}or,\xa0{tex}7 \\alpha = \\frac { 14 } { k } \\text { or } \\alpha = \\frac { 2 } { k }{/tex} ...........(i)Product of roots\xa0{tex}\\alpha ( 6 \\alpha ) = \\frac { 8 } { k }{/tex}or,\xa0{tex}6 \\alpha ^ { 2 } = \\frac { 8 } { k }{/tex}..........(ii)Solving (i) and (ii),{tex}6 \\left( \\frac { 2 } { k } \\right) ^ { 2 } = \\frac { 8 } { k }{/tex}{tex}6 \\times \\frac { 4 } { k ^ { 2 } } = \\frac { 8 } { k }{/tex}or,\xa0{tex}\\frac { 3 } { k ^ { 2 } } = \\frac { 1 } { k }{/tex}or, 3k = k2or, 3k - k2\xa0= 0k[3-k] = 0{tex}\\therefore{/tex}\xa0k = 0 or k = 3k = 0 is not possibleHence, k = 3 | |
| 26636. |
I want a true partner... For long time and with sweet heart |
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| 26637. |
Find the circumference of the circle whose area 16 time the area of the circle with diameter 7 cm |
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| 26638. |
Watch is the sin by cos |
| Answer» Tan... Almost sweating | |
| 26639. |
Find the mean of data using an empirical formula ,when it is given that mode 50.5 and median =45.5. |
| Answer» Mode {tex}= 50.5{/tex}Median {tex}= 45.5{/tex}{tex}3Median = Mode + 2 Mean{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}3 \\times 45.5 {/tex}= 50.5 + 2 Mean{tex}\\Rightarrow{/tex}Mean = {tex}\\frac { 136.5 - 50.5 } { 2 }{/tex}Hence, Mean {tex}= 43{/tex} | |
| 26640. |
Find the zeros of x square +7x + 12 by completing the square method |
| Answer» So easy ?? | |
| 26641. |
4K Jaunga Ek Sath Hai jungle mein dekha Jata Hai Jhanki Bhagwan ka gana Baja de par chala gaya |
| Answer» | |
| 26642. |
When are 10 CBSE exam |
| Answer» Today is releasing the date of CBSE class 10th examination so wait for 3 hour | |
| 26643. |
If you want to chat with your friends pls use WhatsApp or Facebook |
| Answer» | |
| 26644. |
Stop misusing this app |
| Answer» | |
| 26645. |
22*4 |
| Answer» 88 | |
| 26646. |
How should i study for trignometry |
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Answer» Note it down for 5-6 times Nd then learn trigonometric degrees ie: value of sin0^,sin 30^ First of all mesmerise values of trigonometric ratios ie: sin cos etc once you are done with it write it down |
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| 26647. |
What is discriminant |
| Answer» D=-b+4ac | |
| 26648. |
Two dice are thrown simultaneously. What is the probability that a sum less than 6 |
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| 26649. |
5x+4y=566x+10y=20Pair of linear equation |
| Answer» Solve krna h | |
| 26650. |
Paper of periodek test 1 school kv bsf aradpur malda |
| Answer» | |