Explore topic-wise InterviewSolutions in Current Affairs.

Let Viru 's present age = x yrs then Sudhir's present age = (5 + 3x) yrs Anil's present age = (5 + 3x)/2 yrs

Ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5 : 6

e.g., (5 + 3x + x)/3(5 + 3x)/2 = 5/6 => 2(5 + 4x)/3(5 + 3x) = 5/6 => 12(5 + 4x) = 15(5 + 3x)=> 4(5 + 4x) = 5(5 + 3x) => 20 + 16x = 25 + 15x => x = 5

Vinu's age is 5 years.

Asystemof linear equations ishomogeneous if all of the constant terms are zero: Ahomogeneous systemis equivalent to a matrix equation of the form. where A is an m × n matrix, x is a column vector with n entries, and 0 is the zero vector with m entries.

A heterogeneous systemis denned as one consisting of two or morehomogeneousbodies.

We know that the range of principal values of tan^-1, cos^-1 and sin^-1 are (−π/2,π/2) , (0,π) and (−π/2,π/2) respectively,

Let tan^-1(1) = θ1. Then, tanθ1 = 1 = tanπ/4 ⇒θ1 = π/4 ∈[0, π].

Let cos^−1(−1/2)=θ2. Then,cos θ2=−1/2 = −cos π/3=cos(π− π/3)=cos 2π/3So θ2 =2π/3∈[0 , π].

Let sin^−1(−1/2)=θ3. Then, sin θ3=−1/2 = −sin π/6 =sin (−π/6)So θ3 =−π/6∈[−π2 , π2].

So, tan^−1(1)+cos^−1(−12)+sin^−1(−12)= π/4+ 2π/3−π/6=3π/4

homogeneous_a homogeneous mixtures has the same uniform appearanceand composition throughout.many homogeneous mixtures are refferd as solutions for

heterogeneous-a heterogeneous mixture consist of visibly different substances or phases.they are not refferd as solutions

Ahomogeneous mixtureis a solution that has equal proportions of its components throughout any givensample. Conversely, aheterogeneous mixturehas components whose proportions vary throughout thesample. "Homogeneous" and "heterogeneous" are not absolute terms but depend on context and the size of thesample.

Homogeneous mixture

1. A mixture in which different constituents are mixed uniformly.2. They cannot have physically distinct parts.

Heterogeneous mixture

1. A mixture in which different constituents are not mixed uniformly.2. They have physically distinct parts.

Ahomogeneous mixtureis a mixture where the components that make up the mixture are uniformly distributed throughout the mixture. The composition of the mixture is the same throughout. There is only onephase of matterobserved in a homogeneous mixture. So, you wouldn't observe both a liquid and a gas or a liquid and asolidin a homogeneous mixture.

Aheterogeneous mixtureis a mixture where the components of the mixture are not uniform or have localized regions with different properties. Different samples from the mixture are not identical to each other. There are always two or more phases in a heterogeneous mixture, where you can identify a region with properties that are distinct from those of another region, even if they are the same state of matter (e.g., liquid, solid).

A homogeneous mixture is a solution that has equal proportions of its components throughout any given sample. Conversely, a heterogeneous mixture has components whose proportions vary throughout the sample. "Homogeneous" and "heterogeneous" are not absolute terms but depend on context and the size of the sample.

b²+49+14b will be the answer for this

From which book have you taken this question? Please tell us so that we can provide you faster answer.

r = 20cm¤ = 90°Area of sector = theta/360 * π * r2

= 90/360 * 22/7 * 20 * 20

= 1/4 * 22/7 * 20 * 20

= 2200/7314.2 cm2

Area of triangle = 1/2 * base * height

= 1/2 * 20 *20 (base and height both are r ) = 200 cm2

Area of minor segment = 314.2 - 200 = 114.2 cm2

major segment = area of circle - 114.2

= π * r2- 114.2

=22/7 * 20 * 20 -114.2

= 1142 .94 cm2

hai

Like if you find it useful

c.8 answer. D. 10 answer

Let as assume to the contrary that 1/√3 is rational number 1/√3= P/Q { where p and Q are co-prime and Q not equal to 0}√3 P =Q .1 √3 = Q/P √3 = Irrational numberQ/P =RationalIrrational not equal to rational this is a contradiction has arisen by the wrong assumption because of our incorrect assumption that 1 / √3 is rational.hence, 1/ √3 is irrational .{proved}

The parabola is (y+2) ^2 = -4(x-1)

A=

The point P has t = -2 as (at^2,2at) gives (-4,4) and on shifting the origin (x-1) = -1

X= -3

And (y+2) = 4y = 2

As PQ is focal chord and t at Q would be +½ (t1,t2 = -1)

Slope of thenormatis –t

Therefore slope of normal at Q would be -½

Area of shaded region =22/7*14*14*40/360-22/7*7*40/360=51.33cm

Multiplying (1) by 4,12x-16y= 72 (3)Multiplying (2) by 3,-12x-18y=30 (4)Adding 3 and 4,-34y=102y=-3x=18-12/3 =2So the value of x is 2 and y is -3.

Answer will be 0all will be cancelled outOption B

answer is zero... so option is B

answer is 0 means option B is right answer

Bhai booster kansa hai Tere pass

Like my answer if you find it useful!

x=1y=52(1)+5=7this is the correct answer your doubt

this is correct or not

x=1y=5z=7 is the right answer

25. 1. X=0,y=7, x=1,y=5, x=-2,y=11.

Here is your answer

12 is the correct answer me to

x=1y=5it is the correct answer

x = 1y=5z=7 is the correct answer me to

1. (1,5),(2,3),(0,7)

please help me in this also....Out of a no. of saras bird, one fourth of the number are moving about in lot one ninth coupled with one fourth as well as 7 times the square root of no. move on a hill of bird remain in vakula tree. What is the total number of birds?