InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 1451. |
15) . =1600 |
|
Answer» If you find this solution helpful, Please like it. |
|
| 1452. |
\left| \begin array c c 1 - & 5 \\ 7 & 2 \end array \right| \cdot z |
| Answer» | |
| 1453. |
\frac { 15 } { 1600 } |
|
Answer» t q hindi me please |
|
| 1454. |
38. Writeis decimal form and say what kind of decimal expansion it has.(C |
|
Answer» come on brainly fast divide 3 by 13 and you will get the answer. If it is repeated it is recurring and if it is not non recurring |
|
| 1455. |
(-5)/16 %2B 9/7 %2B 7/12 %2B 2/7 %2B 1/8 %2B 5/12 |
|
Answer» 1/8+5/12+2/7+7/12/9/7+ -5/16= (1/8+ -5/16)+(2/7+9/7)+(5/12+7/12)=(2-5)/16 +(11/7)+(12/12)-3/16+11/7+1=-0.1 875+1.57+1=2.38 1÷8+5÷12+2÷7+7÷12+9÷7−5÷16 |
|
| 1456. |
-a^3 %2B 7*a^2 - 5*a %2B a^3 - 5*a^2 %2B 7 |
| Answer» | |
| 1457. |
Letf(x) x and g(x) 2x + 1 be two real functions. Findd g(x) = 2x + l be two real functions·Find |
| Answer» | |
| 1458. |
-1/2 %2B 7/5 %2B 10/2 |
|
Answer» the correct answer is 13/20 |
|
| 1459. |
(d) Speed = 700 m/sec, Time = 14 min |
|
Answer» speed - 700 m pstime - 14 m -14 x 60 -840total distance - 588000 speed =700m/stime =14mminute convert second 1m=60stime = 60×14 =840sDistance = speed × time = 700×840 = 588000mmeter convert km 588000÷1000 = 588km Total distance=588km 58800is the right answer correct answer is 588 km speed 700 mpstime 14m distance.0588000 588000m is correct answer for ur questions speed into time take 1000sec speed=700m/stime=14mminute convert secondtime=60×14=840sdistance=speed×time=700×840=588000mmeter convert km588000÷1000=588km. answer. 588Km . Is the right answer for this questions... The displacement is 50 m. total distance = 588 km speed-700 m pstime-14 m -14×60 -840total distance -588000 Total distance=588 km distance =588 km is the correct answer Given that speed=700 m/sTime=14min=14×60 s. distance=?so,we know that distance=speed ×Time = 700×14×60 m = 588000m = 588000/1000 Km = 588 Km 58800m will be covered in 14 min 588 km is my answer this is correct I am agree with s.p. hi 1+1=11 ok you understand Speed= 700m/secTime= 14 minDistance= speed × time = 700×(14×60) = 700×840 = 588000m =588000/1000 Distance= 588 km so speed will be in 40 minutes equal to42000m per min means 42 km per minand in 14 min the distance will be cover 588000m means 588 km in 14 mins. Total distance =588Km Distance=speed x time =700×14=9800 meter S= 700 m/st = 14 minutes =14×60 second speed=distance/time Distance= speed× time =700 × 14×60 =588000m =588000/1000 km =588 km. Ans... Speed - 700 m/sectime - 14×60 -840Distance - speed× time 700×840 - 588000 m |
|
| 1460. |
1/(2*sqrt(2) %2B 3) %2B 1/(sqrt(7) %2B 2*sqrt(2)) %2B 1/(sqrt(6) %2B sqrt(7)) %2B 1/(sqrt(5) %2B sqrt(6)) %2B 1/(2 %2B sqrt(5)) %2B 1/(sqrt(3) %2B 2) %2B 1/(sqrt(2) %2B sqrt(3)) %2B 1/2 |
| Answer» | |
| 1461. |
If / : R _ {0} → R IS defined by(x)-i-X3, then show that j (x) + J (17 x ) =Prove that the real valued function /(x)=++1 İsaneven funcitonon Rex 12 |
| Answer» | |
| 1462. |
23.Find the domain and range of the real valued function f(x)- |
| Answer» | |
| 1463. |
4ORefind the largest possible domain fo the real valued function t0) |
|
Answer» I think it's Zero...I'm not sure |
|
| 1464. |
5: Find the domain of the real valued functiovariable given by f(a)2-x |
|
Answer» domain is all real Number except 2 where it become undefined wrong see range is -1 and at 2 f(x) is form in 0/0 form. which is undefined |
|
| 1465. |
everLuning or non-terminating repeuning.EXERCISE 1.41. Without actually performing the long division, state whether the following rationalnumbers will have a terminating decimal expansion or a non-terminating repeating decimalexpansion1364ve43125,(1) 455HT1516001000129(vii) 22575() 1577ÂŽ 210 |
|
Answer» 1.Terminating2.Terminating3.Non-Terminating Repeating4.Terminating5.Non-Terminating Repeating6.Terminating7.Non-Terminating Repeating8.Terminating9.Terminating10.Non-Terminating Repeating if denominator is divisible by 2 and 5, the rational number is terminating |
|
| 1466. |
If fx)Vx(x 20)functions, then find fog and gof and check whetherfog gof.and glx) 1 are two real |
| Answer» | |
| 1467. |
Find the domain and range of the real valued function f(x) given by f(x)=4-XX-4Find the domain and range of the function f= (x: +:XER,X+=1} |
|
Answer» 12 divided by 3 exchange driver 2 x is equal to minus 5 x is equal to 9 6 8 7 into 5 into 300 is equal to answer come in 560 R-1/2 is the best answer |
|
| 1468. |
14. i) Let N be the set of natural numbers. Define a real valued functionfiN- Nby fx) 2x+1. Using this definition, complete the table given below267y=f(x) |
|
Answer» from left to right, value of y= f(x) are in order 3,5,7,9,11,13,15 |
|
| 1469. |
iumber is 0.150150015000150000.EXERCISE 1.3Write the following in decimal form and say what kind chas :36100(iii) 4- |
| Answer» | |
| 1470. |
EXERCISE 1.4ctually performing the long division, state whether the following rational1. Without actually performing the long divisionmbers will have a terminating decimal expansion or a non-terminating repeating decimalexpansion:1364160015(iii) 455(1) 3125 |
|
Answer» the best answer me please 2) 17/8multiply by 25 in numerator and denominator17×25/8×25= 425/200= 122.5/100=1.225 it is terminating after 3 decimal place. 2. 17/8 is the correct answer. |
|
| 1471. |
Miscellaneous Exercise on Chapter 10Write down a unit vector in XY-plane, making an angle of 30° with the positnedirection of x-axis. |
| Answer» | |
| 1472. |
An example of such a number is 0.1501ăĽ001EXERCISE 1.31. Write the following in decimal form and say what kind of decima(iii)(vi)has:1) 100(iv) 13362 |
| Answer» | |
| 1473. |
Write where the rational number 7/15 will have a terminating decimal expansion or a non-terminating repeating decimal expansion. |
|
Answer» Since 7/15 have 15 as its denominator and factorization of 15 is 5x3, which is not in the form of 2^m×5^n. Therefore it will have non terminating repeating decimal expansion. |
|
| 1474. |
write whether the rational number 7/15 will have a terminating decimal extension or a nor-lerminating repeating decimal expansion. |
|
Answer» After dividing this we'll get .4666666...Therefore it is non terminating |
|
| 1475. |
(2/5)/14 %2B (3/2)*((2/5)*((-1/6)*((-3)/7))) |
|
Answer» your question is wrong because you don't put the equal sign there is no equal sign please see the question again there is no equal sign |
|
| 1476. |
\lim _ { x \rightarrow \pi } \left( x - \frac { 22 } { 7 } \right) |
| Answer» | |
| 1477. |
In general in an alternating current circuit.(a) The average value of current is 20w.(b) The average value of square of current is zero.(c) Average power dissipation is zero.(d) The phase difference between voltage and current is zero. |
|
Answer» correct option is (a) the average value of current is 20w |
|
| 1478. |
\left. \begin{array} { l } { 1 \frac { 2 } { 15 } \div 3 \frac { 2 } { 5 } - 1 \frac { 1 } { 2 } \text { of } \frac { 1 } { 3 } + \frac { 9 } { 22 } \times ( \frac { 2 } { 15 } + 1 \frac { 1 } { 3 } ) } \\ { 4 \frac { 1 } { 6 } - 2 \frac { 1 } { 3 } \div 3 \frac { 1 } { 2 } \text { of } \frac { 1 } { 3 } + ( 3 \frac { 4 } { 9 } + \frac { 2 } { 3 } ) \text { of } \frac { 3 } { 74 } } \end{array} \right. |
| Answer» | |
| 1479. |
\frac { 3 } { 7 } + ( \frac { - 6 } { 11 } ) + ( \frac { - 8 } { 21 } ) + ( \frac { 5 } { 22 } ) |
| Answer» | |
| 1480. |
3. Evaluate the following.\begin{array}{l}{\frac{-3}{7}+\frac{5}{7}+\frac{1}{7}} \\ {\frac{1}{3}+\frac{7}{8}+\frac{-3}{4}} \\ {\frac{-11}{-4}+\frac{7}{-12}+\frac{-5}{14}+\frac{22}{7}}\end{array} |
|
Answer» thnx . |
|
| 1481. |
\frac{4 \frac{1}{7}-2 \frac{1}{4}}{3 \frac{1}{2}+1 \frac{1}{7}} \div \frac{1}{2+\frac{1}{2+\frac{1}{5-\frac{1}{5}}}}=? |
| Answer» | |
| 1482. |
Using appropriate properties find.(i) -\frac{2}{3} \times \frac{3}{5}+\frac{5}{2}-\frac{3}{5} \times \frac{1}{6}(ii) $ \frac{2}{5} \times\left(-\frac{3}{7}\right)-\frac{1}{6} \times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5} $ |
| Answer» | |
| 1483. |
\frac { 3 } { 4 } \div 2 \frac { 1 } { 4 } \text { of } \frac { 2 } { 3 } - \frac { \frac { 1 } { 2 } } { \frac { 1 } { 2 } + \frac { 1 } { 3 } } \times 3 \frac { 1 } { 3 } + \frac { 5 } { 6 } = ? |
|
Answer» by BODMAS rule3/4÷9/4of2/3-(1/6)/(5/6)×10/3+5/6=3/4÷3/2-1/5×10/3+5/6=1/2-2/3+5/6=(3-4)/6+5/6=-1/6+5/6=4/6=2/3 |
|
| 1484. |
\frac 1 12 %2B \frac 1 20 %2B \frac 1 30 %2B \frac 1 42 %2B \frac 1 56 %2B \frac 1 72 %2B \frac 1 90 %2B \frac 1 110 %2B \frac 1 132 %2B \frac 1 156 = ? |
|
Answer» It can be written as 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + /8*9 + 1/9*10 + 1/10*11 + 1/11*12 + 1/12*13 It can be further expressed as 1/1 - 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + 1/5 - 1/6 + 1/6 – 1/7 + 1/7 - 1/8 + 1/8 - 1/9 + 1/9 – 1/10 + 1/10 – 1/11 + 1/11 – 1/12 + 1/12 - 1/13 Here all the terms get cancelled except the first and last one 1/1–1/13 12/13 Hence 12/13 is the answer Like my answer if you find it useful! |
|
| 1485. |
\frac { 1 } { 9 } , \frac { 2 } { 7 } , \frac { 6 } { 5 } , \frac { 11 } { 12 } , \frac { 7 } { 3 } , \frac { 1 } { 3 } , \frac { 4 } { 7 } , \frac { 8 } { 5 } |
|
Answer» in which class u read class 1 |
|
| 1486. |
In an A.P. the third term is four times the first term, and the sixth term is 17; find the series. |
|
Answer» A3 = 4 a & A6 =17 a+ 2d = 4a3a -2d=0 a =2d/3 A 6=17 a+ 5d =17 2d/3 + 5d =17 17d = 51 d=3 a+ 5d =17 a +15=17 a=2 now series is as follows 2,5,8,11,14,..... |
|
| 1487. |
term can be obtained by34. Find the sixth term of the sequence, whose first term is 1 and (n + 1)adding n to the nth term. |
|
Answer» As given,First term a = 1, (n + 1)th = nth term + n(n + 1)th - nth term = n So,Common difference d = n Therefore,6th term of seriesa6 = a + 5d = 1 + 6n |
|
| 1488. |
.If the sixth term of an AP is zero then show that its 33rd term is threetimes its 15th term.CBSE 2017] |
|
Answer» 6th term = a + 5d = 0 a = -5d33rd term = a + 32d= (-5d) + 32d= -5d + 32d= 27d15th term = a + 14d = (-5d) + 14d= -5d + 14 d= 9d 33rd term = 27d15th term = 9dWhen compared,27d = 3*9dThis means33rd term = 3* 15th term Like my answer if you find it useful! |
|
| 1489. |
The 19th term of an A.P. is equal to three times its sixth term. If its 9thterm is 19, find the A.P. |
| Answer» | |
| 1490. |
The 19th term of an A.P. is equal to three times its sixth term. If its 9th4term is 19, find the A.P |
| Answer» | |
| 1491. |
find the terms14. If 6 times the sixth term of an arithmetic progression is equal totimes the 9th term, find the 15th term |
| Answer» | |
| 1492. |
,11a . he28.times its 15th term.If the sixth term of an AP is zero then show that its 33rd term is three(CBSE 20171 |
|
Answer» A+5d=0 then 33 term means a+32d =a+5d+27d=27d since a+5d=0, then 3 times Of 15 term means 3×(a+14d) =3a+42d= 3a+15d+27d =27d=33rd term Hence proved. Like my answer if you find it useful! |
|
| 1493. |
3 \frac 1 3 \div 6 \frac 3 7 \times 1 \frac 1 2 \times \frac 22 7 = ? |
| Answer» | |
| 1494. |
TILLSUCCO nel candide 332385 votesIl weigh al a boge = UK 9 5009boce that be local in80R uko 8000 500180lo y yooo at 500930100045004 500300179b086egs.to pounded of 4 bed numder-4000b7289Rounded off numder=7300© 8074 d 196297A 8100 A 14600 |
|
Answer» a)3990b)7290c) 8100d)146250 |
|
| 1495. |
A(2, 3), B4, k) आर ((6, - 3) सरेख B)1 |
| Answer» | |
| 1496. |
ा ही न 2,7,12, ८... का 10 पी फ्रव” _क' _. ला _ ले R 2 |
|
Answer» a= 2d= 510th term= दसवां पद होगाa+(10-1)da+9d2+9*5= 47 |
|
| 1497. |
\frac { \sqrt { 7 } - 1 } { \sqrt { 7 } + 1 } - \frac { \sqrt { 7 } + 1 } { \sqrt { 7 } - 1 } = a + b \sqrt { 7 } |
| Answer» | |
| 1498. |
.If a and b are positive integers such that a2 - b4 = 2009, find a + b. |
|
Answer» a² - b⁴ = (a + b²)(a - b²) Now find factors of 2009.Factors of 2009 are = 1, 7, 41,49, 287,2009 Now, we have to use the difference of squares of factorization to obtain (a + b²)(a - b²) = 2009 The prime factorization of 2009 is 7²*41. If we choose two factors 'u' and 'v' such that uv = 2009, a + b² = u and a - b² = v, then 2b² = u-v. If u = 2009, then v = 1 and 2b² = 2008, then, b² = 2008/2 b = √1004 which is not and integer. Now, if u = 287, then v= 7 and 2b² = 280 then, b² = 280/2 b = √140 which is also not an integer. Now, if u = 49, then v= 41 and 2b² = 8 then b² = 8/2 b = √4 b = 2 ……(1) Now substitute value of b =2 in a² - b⁴ = 2009 a² - 2⁴ = 2009 a² - 16 = 2009 a² = 2009 + 16 a² = 2025 a = √2025 a = 45…..(2) So a+b=45+2=47. |
|
| 1499. |
| दि क 3 (050 न A v\_//%‘//// Q = 3° |
|
Answer» sin = x To cos = √(1 - sin²) cos = √(1 - x²) |
|
| 1500. |
7. The distance between two points is 5. One of them is (3, 2) and the ordinate of thesecond is-1 then its x coordinates are1) 7, 12)-7, 13)-7, -14) 7, 1 |
|
Answer» The distance between two points is 5 units. One of the points is (3, 2). The other point has coordinates (x, -1). Use the distance formula to find all possible values of x.d^2 = diffx^2 + diffy^225 = (x-3)^2 + 9x^2 - 6x + 9 + 9 = 25x^2 - 6x - 7 = 0(x-7)*(x+1) = 0x = 7x = -1 |
|