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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.
| 16101. |
(√n×n^5÷3 ) ÷(n^2÷5) |
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| 16102. |
The 8th term of ap is 0 prome that 38th term is triple than 18th term |
| Answer» a(8) = a+7d, where a is the first term and d is the common difference.\xa0a+7d=0\xa0a=-7d\xa0A +7d= 0a= -7dAs is the 38th term\xa0a₃₈ = a +37d= -7d + 37d\xa0= 30da(38) = a+37d = -7d+37d = 30d\xa0a(18)=a+17d = -7d+17d = 10d\xa030 d = 3 times 10 d\xa038th term is triple its 18th term | |
| 16103. |
Find the smallest value of k for real roots xsqure-kx+9 |
| Answer» x2 - kx + 9 = ax2 + bx + ca = 1 . b = -K and c = 9since the eqn has real roots∴ b²-4ac=0⇒ (-k)²-4×1×9=0⇒k²=√36⇒k=±6∵the smallest value of k is required∴k= -6 | |
| 16104. |
How to solve x² +5x+6 by paper cutting and pasting |
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| 16105. |
Paper pattern 2019 2020 |
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Answer» MCQ type questions will be there in the board exam Plz searh the googl ☺️☺️ |
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| 16106. |
Ncert 8.4 |
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Answer» Check on necrt solutions in my cbseguide app under mathematics dashboard What you means this question??? |
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| 16107. |
EvaluateSin 18°/cos 72° |
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Answer» Sin (90-72) / cos72Cos 72/cos721 Sin 18°/cos 72°= Sin18°/cos (90° - 18° )\xa0= sin18°/sin18°\xa0= 1\xa0 1 |
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| 16108. |
Find the sum to n term of the ap 5,2,-1,-4,-7••• |
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Answer» AP = 5 , 2 , -1 , -4 .....Here , First term (A) = 5 and Common Difference = 2-5 = -3An = A + (N-1) × DAn = 5 + ( N -1) × -3An = 5 - 3N +3An = 8+3N Give me my answer |
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| 16109. |
In obejective type mcq will be ask |
| Answer» Yes | |
| 16110. |
Solve the following pair of linear equationsax+by=cbx+ay=1+c |
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| 16111. |
Is there two levels for class 10 math? Easy and hard |
| Answer» Ya maths is divided into to categories standard and basic.if want to take maths in 11 then choose standard other wise choose basic | |
| 16112. |
difference between advance and basic math |
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Answer» Will necrt questions would be asked in the board exam I want to continue with it but aftet exam your decision changes that i want to study maths then while the exam of compartment give the exam of standard In Advance math paper come hard in board exam and. In class 11 you take math subject but basicmath are diff.. In this paper are come easy in. Exam and you not take math..in.11. |
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| 16113. |
prove that cot A +tan A=secAcosecA |
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Answer» thank you LHS :cosA/sinA + sinA/cosA = (cos^2A + sin^2A) /sinA. CosA=1/sinA.cisANow RHS:(1/cosA)(1/sinA)=1/sinA.cosA |
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| 16114. |
Find the area of a rhombus if its vertices are (3,0)(4,5)(-1,4)(-2,-1) taken in order |
| Answer» First divide the rhombus using diagonals and then using coordinates find area of both triangle and then add themm | |
| 16115. |
If angle A and angle P are acute angles such that tanA = tanP .then show that angle A = angle P |
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| 16116. |
2.3×35566÷466 |
| Answer» 175.5403433476 | |
| 16117. |
Circle best questions for class10 |
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| 16118. |
What is the main difference between congruent and similar triangle |
| Answer» in congruent triangle areas are also equal but in similar triangle area are not equal | |
| 16119. |
If x=a sin and y=b tan then prove that a square upon x square minus b square upon y square =1 |
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| 16120. |
Example 15 of trignometry 8.3 |
| Answer» It is from ncert or rd sharma or from where | |
| 16121. |
What is trangle |
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Answer» Traingle is a closed figures made up of three line segments. Triangle is a closed figure formed by adjoining the three sides |
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| 16122. |
Difference Between basic and advanced mathematics |
| Answer» Mathematics – Standard: For the existing level of examination.Mathematics – Basic: For the easier level of examinationThe Standard level will be meant for students who wish to opt for Mathematics at Senior Secondary level and the Basic level would be for students not keen to pursue Mathematics at higher levels. | |
| 16123. |
Chapter:- 12 ex:- 12.3 question no. 13 |
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Answer» I want solution Yes 228 See it on ncert solution |
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| 16124. |
Prove that tan A + sec a - 1 / 10 a minus sec a + 1 is equals to 1 + sin a divided by Cos A |
| Answer» {tex}\\frac{{\\tan A + \\sec A - 1}}{{\\tan A - \\sec A + 1}} = \\frac{{\\tan A + \\sec A - ({{\\sec }^2}A - {{\\tan }^2}A)}}{{\\tan A - \\sec A + 1}}{/tex}\xa0[{tex}\\because{/tex}\xa01 + tan2A = sec2A\xa0{tex}\\Rightarrow{/tex}\xa0sec2A - tan2A = 1]{tex} = \\frac{{\\tan A + \\sec A - \\{ (\\sec A - \\tan A)(\\sec A + \\tan A)\\} }}{{\\tan A - \\sec A + 1}}{/tex}Now, take (tanA + secA) as a common term, we get{tex} = \\frac{{(\\tan A + \\sec A)(1 - \\sec A + \\tan A)}}{{\\tan A - \\sec A + 1}}{/tex}= tanA + secA{tex} = \\frac{{\\sin A}}{{\\cos A}} + \\frac{1}{{\\cos A}}{/tex}{tex} = \\frac{{1 + \\sin A}}{{\\cos A}}{/tex}{tex}\\therefore{/tex} Hence proved | |
| 16125. |
Prove that 5+2 root 3 is an irrational no . |
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Answer» Let us assume that 5+2√3 is rational5+2√3 = p/q ( where p and q are co prime)2√3 = p/q-52√3 = p-5q/q√3 = p-5q/2qnow p , 5 , 2 and q are integers∴ p-5q/2q is rational∴ √3 is rationalbut we know that √3 is irrational . This is a contradiction which has arisen due to our wrong assumption.∴ 5+2√3 is irrational\xa0 See in maths book It\'s not easy to type the answer here man ....yo can easily check it out in your book chapter 1 exercise 2 solved question or you can also watch it on YouTube .... |
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| 16126. |
Name the type of triangle formed by A(_5,6) B(_4,2) C(_1,_5) |
| Answer» It is an isosceles triangle | |
| 16127. |
Aj mera maths ka test h caoching evening me mujhe dar lag rha h....... |
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Answer» All the best Best of luck ???? Best of luck yaar Best of luck..... ???? |
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| 16128. |
Sin 3A. Cos(A-26). Where 3A andA-26 are acute angle |
| Answer» We know that,Sin(90-theta) =cos thetaSo,cos(90-3A) can be compared to cos(90-A-26)90-3A=64-A90-64=-A+3A26=2AA=13°from here onwards u can find the value of 3A and vice versa | |
| 16129. |
Hey is every student free to opt for standard or basic according to his own will |
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Answer» Yes???? Yess YES Yes ..If u want to take maths in ur higher classes then u will choose standard option |
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| 16130. |
Under root 3 sin theta + cos theta = zero find the value of theta |
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| 16131. |
How to solve under root question s |
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| 16132. |
What do you mean by unique sollution |
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Answer» Unique solution means that there is only one solution of the given equation. Unique solution is that in which there is only one solution in a given equation..Hope it will help ?? |
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| 16133. |
Explain theorem 6.3 |
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| 16134. |
Cos(4x-1)=0 solve it by trigonometry |
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Answer» Cos (4x-1)=0Cos(4x-1)=cos904x-1=904x=90+14x=91X=91/4 X=91/4 |
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| 16135. |
How standart maths is different from basic matha |
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Answer» I can wait for my answer till 8:45p.m. I can take maths if I choose basic Bt my teacher told me that I will take maths it i choose basic ...............basic maths is easy than the standard my teacher told me like that Hou is it possible that the level of standard maths is same as the level of previous year questions❓................explain me If i choose basic so I have to give compartment ??????? what is compartment If a person choses basic then he/she has to give a compartment exam in july of standard level to take maths in class 11 The level of question of standard mathematics is same as the level of previous year question The label of basic level math is little bit easier than the standard level maths the only difference is those who choose standard level math will be able to carry math in his next class I mean in class 11th but those who choose basic level mathematics will not able to carry it in class 11 that\'s it The level of a standard label math question is same as the previous year question paper of mathematics |
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| 16136. |
Formulas of ch 8. |
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Answer» It is also Given in summary of ncert book Hey dear u can see in this app in maths content....Thanks @diti? |
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| 16137. |
Scoring chapters of SA1 exams |
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Answer» Trigno,coordinate geometry,AP etc.. Surface area & volume / AP /both Tigno chapter |
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| 16138. |
Find the value of k, if (k+4)xsquare +(k+1) +=0 has equal roots |
| Answer» We have, (k+4) x2 + (k+1)x + 1= 0Here a = (k+4), b = (k+1), c =1{tex}\\implies{/tex}D = b2 -4ac = (k+1)2 - 4 (k+4) (1)= k2 +1 + 2k - 4k -16= k2 -2k -15For equal roots, D = 0{tex}\\implies{/tex}k2 - 2k -15 = 0{tex}\\implies{/tex}k2 - 5k + 3k -15 = 0{tex}\\implies{/tex}k (k - 5) +3 (k - 5) = 0{tex}\\implies{/tex}(k+3) (k-5) = 0Either k+3 = 0 or k-5 = 0{tex}\\implies{/tex}k = -3, 5 | |
| 16139. |
If sin (A-B) =1/2,cos(A+B)=1/2,0 degree |
| Answer» Sin(A-B) =sin30 (we know sin30= 1/2)Cos(A+B)=cos60 (we know cos60=1/2)A-B=30 _(1)A+B=60 _(2) now 2A=90A=45 Put value of A in eq 1 A-B=30 ;B=15 | |
| 16140. |
9sec A-9tanA |
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| 16141. |
2×2×5÷5^6+12 |
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| 16142. |
Taxi fare he cos for 1 km is 15 km and 2 km star 8 km find ap |
| Answer» Sorry but i am not understanding you .so,ask clearly | |
| 16143. |
If ten c =1/cot cThen what is cos c |
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Answer» Sin c /tan c =cos c or 1/ sec c or cos c = cot c ×sin c cos c = sin c / tan c OR cos c = cot c . sin c Sin c× cot c?????? |
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| 16144. |
EvaluateSin18Cos72 |
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Answer» =sin18°cos(90°-18°)=Sin18°sin18°=2sin18° Sin18°Cos 72°/ Sin18°Cos (90°-18°)/ Sin18°Sin18°/ Sin2 18° Sin18°Cos72°Sin18°Cos(90°-18°)Sin18°Sin18°Sin2 18° |
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| 16145. |
Determine if the points (1,5),(2,3)and (-2,-11) are collinear |
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Answer» Answer please |
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| 16146. |
Kya hum basic maths leke engineering krr skte h |
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Answer» No No never . You should take standard for engineering sector No No No Noo |
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| 16147. |
Paragressio with equal common different are known as |
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| 16148. |
Write the sum of first n natural numbers |
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Answer» The sum of n natural numbers is as follows:Sn=n/2(2a+(n-1)d) n(n+1)/2 |
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| 16149. |
Plzz answer jldi do ??? |
| Answer» Hii | |
| 16150. |
If p,q are two prime numbers then what is the HCF and LCM of p and q |
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Answer» HCF will be 1 LCM will be the product of p and q HCF of p and q is 1 and LCM of p and q is pq The “common factor” of p and q is one.Solution: HCF: The full form of HCF is “Highest Common Factor”. The largest common factor of two or more natural numbers is called as “highest common factor”. HCF is also known as “greatest common Divisor (GCD)”.Given that P and q are two prime numbersThe HCF of two p and q prime numbers is one and itself.Here the “common factor” is 1.Therefore, the “common factor” of p and q is one. HCF = 1 and its LCM = pq ok HCF=1 and LCM=pq |
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