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| 64801. |
4. Divide the first polynomial by the secondpolynomial and write the quotient and theremainder.(i) a’ – b3; a - b.(3 marks) |
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Answer» a2-b2 it is the correct answer if this question divide the first polynomial by the second polynomial and write the quotient and the remaider a2-b2 is the correct answer of this question a2-b2 is the correct answer a2--b2 it is the correct answer if this question a2-b2 is the answer of the question a²-b² is tthe correct answer guys like my answer please guys support me plz...plz..plz.. a^3 -b^3 = (a-b)(a^2 +ab+b^2hence (a^3 -b^3)/(a-b)=(a-b)(a^2 +ab+b^2)/(a-b)=(a^2 +ab+b^2)= a³-b³=(a-b)(a²+ab+b²) a3-b3=(a-b) (a2+ab+b2) A2 - b2 .............. a2-b2 is the correct answer of this question a2-b2 is the correct answer if this question a2-b2 it is the correct answer a2_b2 is very correct answer a2-b2 is the correct answer =(a³-b³)\(a-b)=(a-b)(a²+ab+b²)\(a-b)=(a²+ab+b²) quotiont is (a²+ab+b²)and remainder is 0 a^2+ab+b^2 100% sure a square + ab + b square à²-b² is the correct answer a^2+b^2+ab is remainder & 1 is quotient a2_ b2 is the correct answer |
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| 64802. |
6 and l0terms is 44. Find A.P.The sum of first six terms of an A.P. is 42. The ratio of its 10th term to its 3erm is 1: 3. Calculate the first and the thirteenth term of the A.P0th[CBSE 2009 (O.D.) |
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Answer» S6 =42 a + 9d 1------------ = -------a + 29d 3 cross multiply we get3a + 27d = a +29 d2a - 2d = 0 ------------------ (1)its given thatsum of first six terms of an AP is 42therefore S6 = n/2 ( 2a + (n-1) d)42 = 6/2 ( 2a + (6-1) d)42 = 3 (2a + 5d )14 = 2a +5d2a +5d = 14 ------------------- (2) solve eq 1 and 22a - 2d = 02a +5d = 14 we getd= 2a = 213th term of AP= a + (n-1)d2+ (13-1) 2= 2+ 24=26 |
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| 64803. |
{ 23 h - 29 k = 98 } { 29 h - 23 k = 110 } |
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| 64804. |
Divide the first polynomial by the secondpolynomial by synthetic division and find theremainder.(i) 4x3 + 5x-10: x-3. |
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| 64805. |
2Q.21 Ify = sin x + cos 2x, then dy2 equals-(A) sin x +4 sin 2x(B) (sin x + 4 cos 2x)(C) sin x + 4 cos 2x(D) of these |
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| 64806. |
2.Give any two real-life examples for congruent shapes.3. If AABCΔFED under the correspondence ABCFED, write all thecorresponding congruent parts of the triangles.IfADEF4.ΔBCA, write the part(s) of ABCA that correspond toii) EF(iii) ZF(iv) DF |
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Answer» 2) We see many congruent shapes in our day to day life:Biscuits of same packSheets of same letter padTennis balls of same brand and rareTwo same mobile phones etc. |
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| 64807. |
Look at the pairs of triangles given below. Are they congruent? If congruent write thecorresponding parts. |
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Answer» (i) AB = ST AC = RS angle CAB = angle TSR By SAS the two triangles are congruent |
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| 64808. |
sin x cos x51.-ax is equal to1-2 sin xcosx(a) sin 2x+ cfr-sin 2x + c(c) sin 2x + c(d) -sin 2x + c2 |
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| 64809. |
cos 2x drsin x+cos x) |
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| 64810. |
1. a. Given that AMLN E AACB, write down all thecorresponding congruent parts of the triangles.b. If PQ = ZY, RP = YX, ZP = LY, thenAPOREASO |
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Answer» If corresponding angles and sides of two triangles are equal to each other, then triangles are congruent. IfΔABCΔABCandΔFEDΔFEDare congruent under the correspondenceABC↔FEDABC↔FED, then the corresponding sides and angles are equal to each other. All the corresponding parts angles of the triangles are, ∠A↔∠F∠B↔∠E∠C↔∠D∠A↔∠F∠B↔∠E∠C↔∠D All the corresponding parts sides of the triangles are, ¯¯¯¯¯¯AB↔¯¯¯¯¯FE¯¯¯¯¯¯BC↔¯¯¯¯¯¯ED¯¯¯¯¯¯CA↔¯¯¯¯¯¯DF xyz is the correct answer |
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| 64811. |
In the figure, the two triangles are congruent. Thecorresponding parts are marked. We can write ITARAT |
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| 64812. |
(i)the triangles are not congruent.What can you say about their perimeters?fAABC and APQR are to be congruent,name one additional pair of correspondingparts. What criterion did you use? |
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| 64813. |
Look at the pairs of triangles given below. Are thecorresponding parts.y congruent?lfcongruent write theB F0)1304 cm4.5 cm0(iii)(iv)7 cm4.5 cmmos gre con |
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| 64814. |
thell imWrite which partsof the two trianglesABC and ADC arecorresponding16.L Ac-LAA |
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Answer» AB- ADBC- CD angle BAC and angle CAD |
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| 64815. |
29.The radĂź of circular ends of a bucket of height 24 cm are 15 cm and 5 am. Find the area of its curvedsurface.DE 20men |
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Answer» Given: Height = 24 cm Radius 1 = 5 cm Radius 2 = 15 cm To find: Curved surface area. Solution: By formula, Curved surface area = π * Length * ( Radius 1 + Radius 2 ) Here, Length = √ ( Height + ( Radius 2 - Radius 1 )^2 ) Substituting, √ ( 24 + ( 15 - 5 )^2 ) Length = 11 cm Hence, 3.14 * 11 * ( 15 + 5 ) Curved surface area = 690 sq.cm |
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| 64816. |
Example 3: Find the 10th term of the AP : 2, 7, 12, .. |
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Answer» First term,a= 2 Common Difference,d= 5 10th term = a+(10-1)d=2+(9×5)=2+45= 47 |
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| 64817. |
The value of\left\{\cos ^{-1} \frac{1}{5 \sqrt{2}}-\sin ^{-1} \frac{4}{\sqrt{17}}\right\}. The value of tancos-Sis equal to:5v2329292929 |
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| 64818. |
Name the property: 29+4747+ 29 |
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Answer» This is commutative property.In mathematics, a binary operation is commutative if changing the order of the operands does not change the result. It is a fundamental property of many binary operations, and many mathematical proofs depend on it. |
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25. 66 cm3 of silver is drawn into a wire 1 mm in diameter. The length of the wire will be |
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| 64820. |
omial is a factor of the second polynomial by dividingCheck in which case the first a factorthe second polynomial by the first polynomialdiyr 3, 2t +33 - 22 -9t- 12 |
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Answer» Since the remainder is zero, the first polynomial is a factor of the second polynomial. |
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| 64821. |
\cos ^ { - 1 } \frac { 4 } { 5 } + \cos ^ { - 1 } \frac { 12 } { 13 } = \cos ^ { - 1 } \frac { 33 } { 65 } |
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| 64822. |
\begin{array}{l}{\text { 27. } \sqrt{\sin 2 x} \cos 2 x} \\ {\frac{\sin x}{1+\cos x}} \\ {\text { 33. } 1-\tan x}\end{array} |
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Answer» Crop only the question that you want a solution for. We will not be able to provide solutions to multiple questions. |
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| 64823. |
\cos ^ { - 1 } \left( \cos \frac { 13 \pi } { 6 } \right) |
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| 64824. |
2The sides of a triangle are 1, (l +1) and 13 cm. If its perimeter is 56 cm, then find the value of l. |
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Answer» Perimeter is sum of all sides=l+l+1+13=562l+14=562l=56-142l=42l=21 |
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| 64825. |
\operatorname { sin } ^ { - 1 } ( \frac { 12 } { 13 } ) + \operatorname { sec } ^ { - 1 } ( \frac { 13 } { x } ) = \frac { \pi } { 2 } \text { then } x = |
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| 64826. |
1. If the area of square plate is 2025 mm2, then find the sidea. 65 mmc. 45 mmb. 55mmd. 35 mm |
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Answer» Area = 2025Sidde = sqrt(2025) = 45mm a² = 2025 a = √2025 a = 45 mm |
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| 64827. |
न क्ण्थ! wo & |
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Answer» Lets assume that√7 is rational number.ie√7=p/q.suppose p/q have common factor thenwe divide by the common factor to get√7 =a/b were a and b are co-prime number.that is a and b have no common factor.√7 =a/b co- prime number√7= a/ba=√7bsquaringa²=7b² .......1a² is divisible by 7a=7csubstituting values in 1(7c)²=7b²49c²=7b²7c²=b²b²=7c²b² is divisible by 7that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.√7 is irrational. |
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| 64828. |
N WOâ˘M NNw0 X |
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Answer» you haven't calculator |
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| 64829. |
corresponding congruent parts l.1? ABC is right-angled at C. If AC- 5 cm and BC 12 cm find the length of ABSECTION C |
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Answer» using Pythagoras theorem, AB²=AC² + BC² AB²=25+144 AB²=169 AB= 13 cm. |
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| 64830. |
12. The probability of guessing the corec f 2t:1suer tw a certain test is 2 Ifthe prohatbliyof not guessing the correct answer to this question is . then lind the valure of12. The probability of guessing the cornectis certai:i test is 1Ifthe probability3301 |
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Answer» E be the event of guessing correct so E' is event of of incorrect . We know, P(E) + P(E') = 1 p/12 + 1/3 = 1 (p + 4)/12 = 1 p + 4 = 12 p = 8 |
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limx>0 x^10-1024/x^5-32 |
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Answer» please write it on a paper and post it x^10 - 1024 = (x^5)^2 - (2^5)^2 = (x^5 - 2^5 )(x^5 + 2^5) so given term = lim x tends to 0 (x^5 - 2^5 )(x^5 + 2^5) /(x^5 - 2^5 ) = lim x tends to 0 (x^5 + 2^5) = 2^5 please write on paper i cannot understand it please |
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| 64832. |
x : x ^ { \operatorname { log } _ { 0 } x + 2 } = 10 ^ { \operatorname { log } _ { 0 } x + 2 } |
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| 64833. |
9x2.2.LeathBaichung's father is 26 years younger thanBaichung's 9 and faltex and 29 yourse dethan Baiching. The pen of the hos of all thethere is 125 years. What is the age of each |one of them?Bals |
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Answer» PLEASE MARK MY ANSWER AS BEST AND GIVE ME SOME LIKES IF IT'S HELPFULLet, baichung's father's age is = x yearsso, baichung's grandfather's age is = x+26baichung's age is = x-29and the sum of the ages is 185So, x+26+x+x-29=186: 3x-3=186: 3x=189: x=31so, baichung's age is = 31-29 = 2baichung's father's age is = 31baichung's grandpa's age is = 31+26=57 |
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24 Monu's father is 26 years younger than Monu's grandfather and 29 years older than MonThe sum of the ages of all the three is 135 years. What is the age of each one of them? |
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| 64835. |
Renu is 26 year younger than her mother. Hermother is 52 years old. (Take Renu's age as y) |
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Answer» renu age is 26 year olds renu's age is 26 years. Let Renu's age be y y = Mother's age - Renu and mother's age difference y = 52-26= 26 Y= 26 Hence Renu is 26 year's old |
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3. Find x and y from the adjoining figure.5836 |
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| 64837. |
) भ्ु wo \{OQ{ o) A 7९16१ (0o |
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Answer» a condition in which there is difficulty in emptying the bowels, usually associated with hardened faeces. |
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| 64838. |
0) (x+4) (x+10) |
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1 1213i x-11) X12 |
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41cos112-1COS 1 + 0 105 65+ CoS 13Cos 1 33 |
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| 64841. |
\operatorname { sin } ^ { - 1 } \frac { 5 } { 13 } + \operatorname { sin } ^ { - 1 } \frac { 16 } { 65 } = \operatorname { cos } ^ { - 1 } \frac { 4 } { 5 } |
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| 64842. |
L 1 cm ofgold is drawn into a wire of 0.1 mm in diameter. Find thelength of the wire in metre. 1 |
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| 64843. |
11. 1 cm3 of gold is drawn into a wire 0.1 mm in diameter. Find the lengthof the wire. |
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| 64844. |
\begin{array}{c}{\sqrt{12 \sqrt{12 \sqrt{12}}} \ldots-x} \\ {x=?}\end{array} |
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Answer» root(12root(12root(12.....)))=xroot(12 × x)=xby taking square both the sides12x=x^2x^2-12x=0x(x-12)=0x=0 or 12 |
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If m times the mth term of an AP is equal to n times thenth term and mヂn, show that its (m+n)th term is zero |
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| 64846. |
Bac-16) mo. In the adjoiningJomingfigure, Aco to a wechangleFind, andfighterDu |
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Answer» x = 42° and y = 48° |
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| 64847. |
DATE-ä¸Fnd the ualur l |
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Answer» Given beta and 1/beta are roots of equation 4x^2 - 2x + (k-4) = 0 We know product of roots of quadratic equation is equal to c/a For given equation c = (k-4), a = 4Thus, beta*1/beta = (k-4)/4(k-4)/4 = 1k-4 = 4k = 8 Value of k is 8 |
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| 64848. |
Bharat's father is 26 years younger than Bharat's grandaarandfather and 29 you gubtractingthan Bharat. The sum of the ages of all the three is 135 yearsof all the three is 135 vears. What is the ageeach one of them?Example |
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Answer» bharat father's age = 46grand father age= 72bharat age{ 17 bharat father's age = 46grand father age= 72bharat age{ 17 |
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| 64849. |
Example 29. The mass of an electron is9.1x10-31 kg. If its kinetic energy is 3.0 x 10-25 J, Calculateits wavelength |
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| 64850. |
and 3 are the wo rools of the qadraiof a and b.c equation a+x +b-0,then fnd he ralas |
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Answer» If x = -2/3, x = 3 are roots, then the eqn will be (x-2/3)(x-(-3))=0⇒ (x-2/3)(x+3) = 0⇒ x²-(2/3)x+3x-2 = 0⇒ 3x²-2x+9x-6=0⇒ 3x²+7x-6=0 is the required eqn. Compare the eqn with ax²+7x+b=0 Therefore, a=3, b= -6 |
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