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43551.

What is the Thales theorem

Answer» ???
Thank you
Thales theorem where corresponding angle are equal and corresponding side is proprotnal
43552.

How to get perfect marks in Mathematics?

Answer» By revising it again and again.
By hard work & practice
43553.

2+2 is

Answer» 4......LOL? ?
10⁶
43554.

Solve 2x-3y+13=0, 3x-2y+12=0

Answer» Firstly Prepare the proper table of at least two solutions for both the equations..Draw the graph of both the equations from table 1 & from table 2 on the same graph paper with same scale of representation.Then find the coordinates of the point of intersection of the graphs.Here , the two lines intersect at points P (-2,3)Hence, x= -2 & y= 3Thus, we have the following table giving points on the line 3y-2y+12 =0Hence, x =-2 ,y=3\xa0is the solution of the given system of equations.
43555.

3x+y-12=0 ,x-3y+6=0 by graphical method

Answer»
43556.

X+y=7,2x-3y=11 by substitute method

Answer» Thanks
(1) Given Equation is x + y = 7 x = 7 - y --------- (1)(2) 2x - 3y = 11 2(7 - y) - 3y = 11 14 - 2y - 3y = 11 14 - 5y = 11 -5y = 11 - 14 -5y = -3 y =\xa0Substitute y =\xa0\xa0in (1), we getx = 7 -\xa0 =\xa0 =\xa0Therefore the value of x =\xa0, y =\xa0
43557.

3sin square 60 degree + sec square 30 degree + 2 sin90 degree + 3 cos 0 degree -tan square 60 degree

Answer»
43558.

If An=5n-2 then find S4

Answer» First find A1Next find A4Now, we have A=A1, L=A4 and N=4By putting these values in formula of Sn=n/2 (a+l) we will get S4= 42
S4=42
43559.

In the given figure, if PQRS is a parallelogram and AB || PS, then prove that OC || SR.

Answer» To prove - OC║SRProof - In ΔOPS and ΔOAB∠POS = ∠AOB (common in both)∠OSP = ∠OBA (corresponding angles are equal as PS║AB)=> ΔOPS ~ ΔOAB [AA criteria]=> PS/AB = OS/OB ........................(1) (sides in similar triangles are proportional)In ΔCAB and ΔCRQAs, QR║AB=> ∠QCR = ∠ACB (common)=> ∠CBA = ∠CRQ (corresponding angles are equal)=> ΔCAB ~ ΔCQR [AA criteria]=> CR/CB = QR/AB (sides in similar triangles are proportional)Also, PS = QR [ PQRS is parallelogram]=> CR/CB = PS/AB ......................(2)From (1) and (2)=> OS/OB = CR/CB=> OB/OS = CB/CRSubtracting 1 from both sidesSo, OB/OS - 1 = CB/CR - 1=> (OB - OS)/OS = (CB - CR)/CR=> BS/OS = BR/CRBy converse of Thales Theorem=> OC║SR . Hence proved .
43560.

Explain why 7×11×13+13 and 7×6×5×4×3×2×1+5 are composite numbers?

Answer» That is why both no. Are composite
Numbers are of two types - prime and composite.Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.It can be observed that=\xa0=\xa0=\xa0The given expression has 6 and 13 as its factors other than 1 and number itself.Therefore, it is a composite number.=\xa0=\xa0=\xa01009 cannot be factorized furtherTherefore, the given expression has 5 and 1009 as its factors other than 1 and number itself.Hence, it is a composite number.
,xdhh
43561.

(cosec A -secA )(cot A-Tan A)=(cosec A+secA )(SecA cosecA -2 )

Answer» LHS:-(cosecA-secA)( cotA-tanA) => (1/sinA-1/cosA)(cosA/sinA-sinA/cosA)=> (cosA-sinA/sinAcosA)(Cos^2A-sin^2A/sinAcosA)=> (cosA-sinA)(cosA+sinA)(cosA-sinA) /(sinAcosA)^2=> (cosA-sinA)^2(cosA+sinA)/(sinAcosA)^2=> (cos^2A+sin^2A-2cosAsinA)(cosA+sinA)/(sinAcosA)^2=> (1-2cosAsinA)/(cosAsinA) (cosA+sinA)/(cosAsinA)(since cos^2A+sin^2A=1)=> (1/cosAsinA-2cosAsinA/cosAsinA)(cosA/cosAsinA+sinA/cosAsinA)=> (secAcosecA-2)(cosecA+secA)Hence, LHS=RHS
43562.

Proove thales formula

Answer» \xa0Basic Proportionality Theorem (Thales theorem):\xa0If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio.\xa0\tIn ∆ABC , if DE || BC and intersects AB in D and AC in E then AD AE ---- = ------ DB EC\tProof on Thales theorem :If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.Given :\xa0In ∆ABC , DE || BC and intersects AB in D and AC in E.Prove that :\xa0AD / DB = AE / ECConstruction :\xa0Join BC,CD and draw EF ┴ BA and DG ┴ CA.\xa0\xa0\t
Statements
Reasons
1) EF ┴ BA1) Construction2) EF is the height of ∆ADE and ∆DBE2) Definition of perpendicular3)Area(∆ADE) = (AD .EF)/23)Area = (Base .height)/24)Area(∆DBE) =(DB.EF)/24) Area = (Base .height)/25)(Area(∆ADE))/(Area(∆DBE)) = AD/DB5) Divide (4) by (5)6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC6) Same as above7) ∆DBE ~∆DEC7) Both the ∆s are on the same base and between the same || lines.\xa08) Area(∆DBE)=area(∆DEC)8) If the two triangles are similar their areas are equal9) AD/DB =AE/EC9) From (5) and (6) and (7)\t
43563.

If tanA=4/5 then find the Value of cosA-sinA/cosA+sinA

Answer» Tan A =4/5=B/CCosA _ SinA /CosA+SinA=4_5/root 41/4+5/root411/9
1/9
43564.

Cosec thetha - sin thetha) ( sec thetha- cos thetha)

Answer» cosA×sinA
43565.

480/x-8 -480/x = 3 write in the form of Quadratic equation.

Answer»
43566.

How many terms of the AP : 9, 17, 25, must be taken to give a sum of 636?

Answer» Let the number of terms required to make the sum of 636 be n and common difference be d.Given Arithmetic Progression : 9 , 17 , 25 ....First term = a = 9Second term = a + d = 17Common difference = d = a + d - a = 17 - 9 = 8From the indentities of arithmetic progressions, we know : -, where\xa0\xa0is the sum of first term to nth term of the AP, a is the first term, d is the common difference and n is the number of terms of AP.In the given Question, sum of APs is 636.Therefore,= > 4n² + 5n - 636 = 0= > 4n² + ( 53 - 48 )n - 636 = 0= > 4n² + 53n - 48n - 636 = 0= > 4n² - 48n + 53n - 636 = 0= > 4n( n - 12 ) + 53( n - 12 ) = 0= > ( n - 12 )( 4n + 53 ) = 0By Zero Product Rule,= > n - 12 = 0= > n = 12Hence,Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.\xa0
43567.

How many three digit number divisible by 7?

Answer» here a= 105l= 994n=?d= 7so994 = 105+(n-1) 7= 105+7n-7994= 98+7n994-98 = 7n896/7 = nn= 128so 128 three digit numbers are divisible by 7
Given,A(first term)=105,An(last term)=994,d( common difference)=7,n=?An=A+(n-1)d995=105+(n-1)7887=(n-1)7n-1=127n=127+1=128Hence, there are total 128 three digit no. divisible by 7.
43568.

(sin A + cosec A)2 + (cos A + sec A )2 = 7 tan2 A +cot2 A

Answer»
43569.

Obtain all other zeroes 3x⁴ + 6x⁴ - 2x² -10x - 5, if two of its zeroes are √5/3 and -√5/3.

Answer» (X-1) ,(x-1),-1,-1
43570.

How is H3O is formed?

Answer» H2O+H^ion=H3O
H2O+H^+ ion =H3O
43571.

2√7=

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43572.

Determine whether the given quadratic equation have real roots so find root xsquare+x+2=0

Answer» No real roots
43573.

Ncert lesson 1 example no. 5.

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43574.

Find mean & mode, medium

Answer» Your answer : Mode-The most repetitive number! -\xa0Median:The number in the MIDDLE when they are IN ORDER! -\xa0Mean- The\xa0AVERAGE\xa0OF ALL NUMBERS: You add up all the numbers then you divide it by the TOTAL NUMBER of NUMBERS!
43575.

The least no. That is divisible by all the no. from 1 to 5(both inclusive) is

Answer» I think 120
43576.

The 5 term and 15 term of an ap is 13 and - 17

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43577.

How many two digit number are divisible by 6 ?

Answer» 15
an=a+(n-1) d96=6+(n-1) 696=6+6n-696=6nn=96/6=16
1
12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90 & 96. HOPE IT HELPS YOU?
15(FIFTEEN). HOPE IT HELPS YOU?✌
43578.

[3^(x+2)]+[3^(-x)]=10

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43579.

For any positive real number \'a\' ,prove that there exists an irrational number b such that 0

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43580.

Prove that.cot A-cos A/cot A+cos A=cosecA-1/cosecA+1

Answer» LHS CotA-cosA/cotA+cosADivide numerator and denominator by cosA=(cotA/cosA-cosA/cosA)/(cotA/cosA+cosA/cosA)=(cosA/sinA•1/cosA-1)/(cosA/sinA•1/cosA+1)(1/sinA-1)/(1/sinA+1)=(cosecA-1)/(cosecA+1)(since 1/sinA=cosecA)=RHS
43581.

Find the area of circle whose cicumferrence is 22cm.

Answer» Circumference of circle = 22 cmLet the radius of the given circle be \'r\'Circumference of a circle = 2πr⇒ 22 = 2\xa0× 22/7\xa0× r⇒ r = (22\xa0× 7)/(22 × 2)⇒ r = 154/44⇒ r = 3.5 cmArea of circle =\xa0πr²⇒ 22/7\xa0× 3.5 × 3.5⇒ 269.5/7= 38.5 cm²So, the area of the given circle is 38.5 cm²
43582.

The sum and product of the zeroes of the polynomial 9xpower2-5 respectively are:

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Sum = 5/9,Product = 0
Ask someone else
I don\'t know
43583.

Factorise x^2-270x -1800000

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43584.

If sec theta is equal to 5.4 so that 2 cos theta minus sin theta upon cos theta minus 10 theta

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43585.

978764

Answer» Bahi aaj to ye message send kar diya ha nest time mat karana
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????
978 764 = 978 764. HOPE IT HELPS YOU?✌?
978 764 is equal to
43586.

767/777

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0.987129987
0.999
43587.

BBB

Answer» Better Business Bureau
What is your question??
??????????
43588.

Hcf of 34 56

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2
2
2
2
43589.

A dice is thrown once . The probability of getting a prime number is?

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The probability of getting a prime number=1/2. HOPE IT HELPS YOU?✌
Possible outcomes when a die is thrown = {1,2,3,4,5,6}Number of possible outcomes once = 6(i) Let E be the event of getting a prime number.Prime numbers on the die are = {2,3,5}Number of favourable outcomes = 3•p(E)=favourable outcomes /total outcomes =3/6=1/2
1/2
43590.

If u wants to study with us firstly give ur no. in this answer box

Answer» Wat?¿? #Ritik Kumar???
6202983023
9304208420
Cheater
9087654321
43591.

cos2A - 3 cos A + 2 / sin 2 A = 1

Answer» A=60°
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43592.

What is main difference between rational and irrational

Answer» Rational not have root . Irrational have roots
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\tS.NoRational NumbersIrrational Numbers1Numbers that can be expressed as a ratio of two number (p/q form) are termed as a rational number.Numbers that cannot be expressed as a ratio of two numbers are termed as an irrational number.2Rational Number includes numbers, which are finite or are recurring in nature.These consist of numbers, which are non-terminating and non-repeating in nature.3Rational Numbers includes perfect squares such as 4, 9, 16, 25, and so onIrrational Numbers includes surds such as √2, √3, √5, √7 and so on.4Both the numerator and denominator are whole numbers, in which the denominator is not equal to zero.Irrational numbers cannot be written in fractional form.5Example: 3/2 = 1.5, 3.6767Example:\xa0√5, √11\t
43593.

2x-3y+10

Answer» Bye
Kaya hai
Algebra hai
Solution karna hai
Kaya karna hai
43594.

Where can I find the deleted content of maths of class 10

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In my cbse guide
43595.

If an AP if a=4,n=7,d=4then nth term will be

Answer» An+(n-1) dAn=4+(7-1) 4An=4+6×4An=4+24An=28
an= 28
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An = 28 by an = a + (n - 1)× d
43596.

Surface areas formulas

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Surface area of a cuboid =2(lb+bh+hl) sq. unitSurface area of a cube = 6s2 sq. unit
43597.

Find the value of (26+15√3)^2/3 -(26+15√3)^-2/3

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43598.

Real life application for section formula

Answer»
43599.

Real life application of section formula

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43600.

Find the LCM and HCF of 12, 15 and 21 by the prime factorization method.

Answer» Given : 12, 15 , 21 Prime factor of 12 : 2*2*3Prime factor of 15 : 3*5Prime factor of 21 : 3*7HCF of 12, 15 ,21 : 3LCM of 12 , 15 , 21 : 420