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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.
| 8951. |
Solve:0.2x+0.3y=1.3, 0.4x+0.5y=2.3 |
| Answer» Given :To Find : solve using substitution methodSolution : -----1\xa0------2Substitute the value of x from 1 in 2So, x = 2 and y = 3 | |
| 8952. |
Show that 13*17*19*23+13*23 is composite number |
| Answer» We will check the result13 × 1 7 × 19 × 23 + 13 × 23221 × 437 + 29996577 + 29996876Let\'s do factorization96876 = 2 × 2× 3 × 3 × 3× 3 ×13 × 23\xa0It is divisible by numbers other than 1 and the number itself.So, it is a composite number\xa0 | |
| 8953. |
Find the sum of all 2 digit odd positive numbers |
| Answer» Two digit odd positive numbers are 11,13,15,17...........99 are in A.P.Here a = 11 and d = 2, tn= 99, n = ?Sum of the n terms = (n/2)[2a+(n -1)d]But tn = a + (n -1)d⇒ 99 = 11+ (n-1)2⇒ 99 -11 = (n-1)2⇒ 88/2 = (n-1)∴ n = 45.subsitute n = 45 in sum of the n terms we obtain⇒ s45 = (45/2)(2×11 + (45 -1)2)⇒ s45 = (45/2)(110)⇒ s45 = 45×55.⇒ s45 = 2475.∴sum of all two digit odd positive numbers = 2475. | |
| 8954. |
What are the deleted portion of maths? |
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Answer» Thanks Ayush ? Plz. Give me ? But Don\'t think that anything is deleted in mathematics https://deleted.cbse.math.2020.2021/j/73791433703?pwd=cWwxSnY2MUpmMUtxbzhndXlHN3hBdz09 |
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| 8955. |
Solve : x/x-1 + x-1/x =85/42 (x≠0,1) |
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Answer» I want written solution... x=7 or -6 I nown |
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| 8956. |
Dividend is equal to |
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Answer» Divisor into quotient+remainder Two water taps together can fill a tank in Hi |
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| 8957. |
divide 24 in three parts such that they are in AP and their product is 440 |
| Answer» The parts are in AP, let the terms be a-d, a and a+dwhere the first term =a-dthe common difference =dGiven, sum =24cma-d + a+a +d = 24cm3a = 24cma =8cmGiven: Product of the terms = 440(a-d) (a) (a+d) = 440(8-d) (8+d) (8) =440(8-d)(8+d) = 5582\xa0-d2\xa0= 5564 - d2\xa0= 55d2\xa0= 64-55d2\xa0= 9d = 3 or -3AP- a-d , a, a+dso the AP is 8-3, 8, 8+3 or 8+3,8,8-3AP = 5, 8, 11 or 11, 8, 5 | |
| 8958. |
How to derive area of triangle formua |
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Answer» Explanation. In the given triangle b is the base of triangle and h is the height of triangle multiply base and height along with haph. Why haph because the reason is a triangle is a polygon with three side . A diagonal of a rectangle separates the rectangle into two congruent triangle. The area of each triangle is 1/2 the area of rectangle. So the area of A of a triangle is given by formula A= 1/2bh where b is the base and h is the height of a triangle. Area of triangle ???= 1/2* base * height |
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| 8959. |
Determine the prime factorization of the number 556920 |
| Answer» \the prime factors are: 2 x 2 x 2 x 3 x 3 x 5 x 7 x 13 x 17\tWritten in exponential form: 23\xa0x 32\xa0x 51\xa0x 71\xa0x 131\xa0x 171\t\xa0556,920\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa02\xa0278,460\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa02\xa0139,230\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa02\xa069,615\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa03\xa023,205\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa03\xa07,735\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa05\xa01,547\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa07\xa0221\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa0\xa013\xa017\t | |
| 8960. |
Find the point on the x- axis which is equidistant from (2,-5) and (-2,9) |
| Answer» Given points\xa0A(2,−5)\xa0and\xa0B(−2,9)Let the points be\xa0P(x,0).So,\xa0AP=PB\xa0and\xa0AP2=PB2\xa0⇒(x−2)2+(0+5)2=(x+2)2+(0−9)2⇒x2+4−4x+25=x2+4+4x+81⇒x2+29−4x=x2+85+4x⇒−4x−4x=85−29⇒−8x=56⇒x=−7Hence, point on the\xa0x-axis which is equidistant from\xa0(2,−5)\xa0and\xa0(−2,9)\xa0is\xa0(−7,0). | |
| 8961. |
Like √3 etc |
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| 8962. |
Some important question about triangle chapter., |
| Answer» If you are in the session of 2020-21 doing only ncert questions | |
| 8963. |
Deffrence between simalarites or congruence |
| Answer» Congruent\xa0means being exactly the same. When two line segments have the same length, they are congruent. When two figures have the same shape and size, they are congruent.\xa0These two triangles are congruent. They are exactly the same in every way. They are the same size and the same shape. Their side lengths are the same and that their angle measures are the same.Sometimes, two figures will be similar. Similar means that the figures have the same shape, but not the same size. Similar figures are not congruent.\xa0These two triangles are similar. They are the same shape, but they are not the same size. | |
| 8964. |
Arun got a playing top |
| Answer» | |
| 8965. |
Sin6A+cos6A=1-3sin2Acos2A |
| Answer» a n s w e r\xa0sin^6A + cos^6A= (sin^2A)3 + (cos^2A)3= (sin^2 A + cos^2 A) (sin^4 A – sin^2A cos^2A + cos^4A)= 1 (sin^4A – sin^2Acos^2A + cos^4A + 2sin^2Acos^2A – 2sin^2Acos^2A)= (sin2A + cos2A) – 3sin2A cos2A= 1 – 3sin^2Acos^2A | |
| 8966. |
How to solve any problem based on trigonometry? |
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Answer» , ?? Hmm |
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| 8967. |
prove that sin A - cos A +1/sin A-cos A-1=1/sec A-tan A |
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| 8968. |
Find the point of trisection of the line segment joining the points (1). (5, -6) |
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| 8969. |
if sin theta =cos theta ,find the value of theta and tan theta |
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Answer» Theta = 45°Tan theta =1 S i n t h e t a = c o s t h e t a\xa0sin theta / cos theta = 1tan theta = 1tan theta = tan 45°theta = 45° |
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| 8970. |
The sum of first n odd natural numbers is |
| Answer» First n natural numbers1 , 3 , 5 , 7 , 9 ... ( 2n - 1 ).Method 1 =>Method 2 | |
| 8971. |
prove that 15+17 root 3 is irrational |
| Answer» Let 15 + 17√3 is a rational number.\xa0p/q = 15 + 17√3 q is not equal to zero\xa0p and q are integers.\xa0p/q = 15 + 17√3\xa0=> p - 15/q = 17√3\xa0Rational number = Irrational number\xa0This is contridication.\xa0Hence , 15 + 17√3 is an irrational number | |
| 8972. |
If tanA =√2 -1 show that tanZ/1+tan^2A=√2=4 |
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| 8973. |
What is alpha beta |
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Answer» a n s w e rA l p h a and beta are the Greek letters that are used in mathematics to denote the constant values such as the roots of polynomials. What is your name |
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| 8974. |
What is rational + irrational |
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Answer» Irrational number Rational Rational form of p/q & q≠0 Irrational Rational numbers are the numbers that can be expressed in the form of a ratio (P/Q & Q≠0) and irrational numbers cannot be expressed as a fraction. But both the numbers are real numbers and can be represented in a number line. |
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| 8975. |
Use maths operation,+, -, /and multiply make answer zero by 8,8,3,3 |
| Answer» 8*3-8*3=24-24=0 | |
| 8976. |
What is log in math |
| Answer» In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b, must be raised, to produce that number x. | |
| 8977. |
Sin 2 B = 2 Sin B is true when B is equal to |
| Answer» The value of B is 0°Step-by-step explanation:given sin 2B = 2 sin Bwe have to find the value of Bthenhence , The value of B is 0° | |
| 8978. |
Prove that 0+0 = 1 |
| Answer» 0+0 = 1 | |
| 8979. |
Why 1/4 and -1 are zeros of the polynomials p(x)=4x²+3x-1? |
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Answer» When we factorize 4x2-4x+x-1 4x(x-1)+(x-1) to give two zeroes because it is quadratic polynomial As when we factorise it 4x2-4x+x-14x(x-1)+1(x-1)x=1,-1/4. |
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| 8980. |
Ba |
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| 8981. |
2√3x2+5x+√3 |
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Answer» Thank you ? 2√3x2 ˗ 5x\xa0+ √3\xa0\xa0⇒ 2√3x2 ˗ 2x\xa0˗ 3x+ √3 ⇒ 2x\xa0(√3x\xa0˗ 1) ˗ √3 (√3x\xa0˗ 1) = 0\xa0\xa0⇒ (√3x\xa0˗ 1) or (2x\xa0− √3) = 0 ⇒ (√3x\xa0˗ 1) = 0 or (2x\xa0− √3) = 0 ⇒ x = 1/ √3 or x = √3/ 2 ⇒ x = 1/ √3 × √3 /√3 = √3/ 3 or x = √3/ 2\xa0 2√3x2+5x+√3 |
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| 8982. |
What is cosec? |
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Answer» Chal Cosec = Hypotenuse /PerpendicularIt\'s just the opposite of sin Photo bhejna hai |
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| 8983. |
.If x= 2 and x+ y 5, then the find the value of y. |
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Answer» Y=3 .If x= 2\xa0x+ y = 5\xa02 + y = 5y = 5 - 2y = 3 |
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| 8984. |
9+2 |
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Answer» 11 11 11 11 11 |
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| 8985. |
If -5 is a root of 2 x²+px-15=0 and p(x²+x) +k=0has equal roots find p and k ? Pls help...... |
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| 8986. |
Tan30degree cosec60degree+tan60degree sec 30degree |
| Answer» 2 | |
| 8987. |
The point a(9,0), b(9,6), c(-9,6) and d(-9,0) verties a |
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Answer» AB= √(9-9)²+(6-0)²\xa0= 6BC= √(-9-9)²+(6-6)²\xa0= 18CD = √(-9+9)²+(0-6)²\xa0= 6DA = √(-9-9)²+(0-0)²\xa0= 18so,it\'s opposite sides are equal,hence it is rectangle. Tell me |
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| 8988. |
At present x=6 and y=3 now if x=70 then find y |
| Answer» At present x=6 and y=3 now if x=70 then find ySolution:- let the X and xx=6 and after exceeding X=70so, exceeding no.= X-x (70-6)= 64Since, x exceed by 64 So, Y= y+64=(3+64)= 67 | |
| 8989. |
If secA=15/7 and A+B=90°. Find the value of cosecB. |
| Answer» Sec A = 15/7 Cos A = 1/SecA Therefore CosA = 1/ 15/7 = 7/15\xa0A + B = 90 or A = 90 - B\xa0CosA = Cos(90-B)\xa07/15 = SinB\xa0CosecB = 1/SinB = 1/ 7/15 = 15/7\xa0 | |
| 8990. |
Remind the AP whose third term is 16 and the 7th term exceeds the 5th term by 12 |
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Answer» Sum of n term balavi board Mein aaega We have given\xa0a3\xa0= 16a\xa0+ (3 − 1)\xa0d\xa0= 16a\xa0+ 2d\xa0= 16 (1)a7\xa0−\xa0a5\xa0= 12[a+ (7 − 1)\xa0d] − [a\xa0+ (5 − 1)\xa0d]= 12(a\xa0+ 6d) − (a\xa0+ 4d) = 122d\xa0= 12d\xa0= 6From equation (1), we obtaina\xa0+ 2 (6) = 16a\xa0+ 12 = 16a\xa0= 4Therefore, A.P. will be4, 10, 16, 22, … |
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| 8991. |
Express each number in a product of its prime factor |
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Answer» Question 1. Express each number as a product of its prime factors:(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429Answer:(i) 140 = 2 x 2 x 5 x 7 =2² x 5 x 7(ii)156 =2 x 2 x 3 x 13 =2² x 3 x 13(iii)3825 =3 x 3 x 5 x 5 x 17 =3² x 5² x 17(iv)5005 =5 x 7 x 11 x 13 =5 x 7 x 11 x 13(v)7429 =17 x 19 x 23 =17 x 19 x 23 Express each number in a product of its prime factors |
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| 8992. |
List out all theorems from triangle ? chapter and prove it |
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| 8993. |
For what value(s) of ‘a’ quadratic equation 30ax^2-6x+1=0 has no real roots? |
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Answer» The equation has no real root i. e. B2-4AC <0here b=-6, a=3a and c=1(-6*-6) -4(3a)(1) <036-12a<036<12a36/12 Here,coefficients of standard form of equation Ax²+Bx+C=0 are A=30a B=-6 C=1 So,Discriminant,D=B²-4AC Since,equation has no real roots i.e. D<0 i.e. B²-4AC<0 i.e. (-6)² -4 (30a)(1)<0 i.e. 36-120a <0 On transposing (-120a) i.e. 36 <0-(-120a) i.e. 36 <120a i.e. 6×6 <6×(20a) On cancelling out 6 from both sides i.e. 6<20a On dividing both sides by 20 i.e. 6/20 < a i.e. 3/10 < a i.e. a >3/10 Hence, a>3/10 for which 30ax² +6x+1=0 gas no real roots B= ±6 as B²=36 Here,coefficients of standard form of equation Ax²+Bx+C=0 areA=30aB=6C=1So,Discriminant,D=B²-4AC Since,equation has no real rootsi.e. D<0i.e. B²-4AC<0i.e. (6)² -4 (30a)(1)<0i.e. 36-120a <0 On transposing (-120a)i.e. 36 <0-(-120a)i.e. 36 <120ai.e. 6×6 <6×(20a)On cancelling out 6 from both sidesi.e. 6<20a On dividing both sides by 20 i.e. 6/20 < a i.e. 3/10 < ai.e. a >3/10 Hence, a>3/10 for which 30ax² +6x+1=0 gas no real roots |
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| 8994. |
Kis ko math ki revision ha dang sa ?????? |
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Answer» Dkh leyo Mujhe? Mujhe ? |
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| 8995. |
Case study based question |
| Answer» You can find in sample paper of (basic or standard) in this app only. | |
| 8996. |
Case study vale questions kaha se milege? |
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| 8997. |
Which term of the ap 3,15,27,39,........will be 120 more than its 21st term ? |
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Answer» Given sequence is\xa03,15,27,39,...The first term is\xa0a=3The common difference is\xa0d=15−3=12we know that the\xa0nth\xa0term of the arithmetic progression is given by\xa0a+(n−1)dLet the\xa0mth\xa0term be\xa0120\xa0more than the\xa021st\xa0termTherefore,\xa0120+mthterm=21stterm⟹120+a+(m−1)d=a+(21−1)d⟹120+(m−1)12=(20)12⟹12m−12=240+120⟹12m=360+12⟹12m=372⟹m= 372/12 \u200b=31 a = 3, d=12a21 = a + (n-1)d= 3 + (21-1)12= 243243 + 120 = 363an = 363 a + (n-1)d = 3633 + (n-1)12 = 36312n = 363 - 3 + 12n = 372 / 12∴n = 31st term |
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| 8998. |
Proof of AAA similarity criteria |
| Answer» AAA similarity theorem or criterion:If the corresponding angles of two triangles are equal, then their corresponding sides are proportional and the triangles are similarIn ΔABC and ΔPQR,\xa0∠A = ∠P , ∠B = ∠Q , and\xa0∠C =\xa0∠R then AB PQ\xa0= BC QR\xa0= ACPRand ΔABC ∼ ΔPQR.Given: In\xa0ΔABC and\xa0ΔPQR,\xa0∠A =\xa0∠P,\xa0∠B =\xa0∠Q,\xa0∠C =\xa0∠R.To prove: AB PQ\xa0= BC QR\xa0= ACPRConstruction : Draw LM such that PL AB\xa0=\xa0PM AC\xa0.Proof: In\xa0ΔABC and\xa0ΔPLM,AB = PL and AC = PM (By Contruction)∠BAC =\xa0∠LPM (Given)∴\xa0ΔABC\xa0≅\xa0ΔPLM (SAS congruence rule)∠B =\xa0∠L (Corresponding angles of congruent triangles)Hence\xa0∠B =\xa0∠Q (Given)∴\xa0∠L = ∠Q\xa0LQ is a transversal to LM and QR.Hence ∠L = ∠Q\xa0(Proved)∴ LM\xa0∥ QR PL LQ\xa0= PM MR\xa0 LQ PL\xa0= MR PM (Taking reciprocals) LQ PL\xa0+ 1 = MR PM\xa0+ 1 (Adding 1 to both sides) LQ+PL PL\xa0= MR+PM PM\xa0 PQ PL\xa0= PR PM PQ AB\xa0= PR AC (AB = PL and AC =PM) AB PQ\xa0= AC PR (Taking Reciprocals) ............... (1) AB PQ\xa0= BC QR AB PQ\xa0= AC PR\xa0= BC QR\xa0∴\xa0ΔABC\xa0~\xa0ΔPQR | |
| 8999. |
The LCM of two numbers is 1200.Which of the following cannot be their HCF?A)600B)500C)400D)200 |
| Answer» Because the\xa0HCF\xa0of\xa0two numbers\xa0is also the factor of\xa0their\xa0multiples, therefore\xa0200,\xa0400\xa0and\xa0600\xa0divide\xa01200\xa0but\xa0500\xa0does not. | |
| 9000. |
The angle of elevation of a ladder learning against a wall |
| Answer» The angle of elevation of a ladder leaning against a wall\xa0is 60º and the foot of the\xa0ladder\xa0is 4.6 m away from the\xa0wall.\xa0Angle\xa0= 60º and distance between\xa0wall\xa0and\xa0ladder= 4.6 m. = 9.2 m.\xa0Angle\xa0is 60 degree, on\xa0angle\xa0will be 90 degree, so 3rd\xa0angle\xa0is 30 degree. | |