Explore topic-wise InterviewSolutions in Current Affairs.

√4x - 3 + √2 x + 3

2x +√2 x

x ( 2 + √2)

Theperiodictable is arranged according toperiodic propertiesin terms of ionization energy, electronegativity, atomic radius, electron affinity, and metallic character. Theperiodictable arranges the elements byperiodic properties, which are recurring trends in physical and chemical characteristics.

What is EN fullform?

The full form of EN is electronegativity

if a and b vary inverse

a= k/b

=> 7 = k/b = k/8 => k = 7*8 = 56

so, when b = 4 => a = 56/4 = 14

when a = 28=> b = k/a = 56/28 = 2

x ,6, 8 ,12 ,18y ,18 ,26,36,63

x ,6,8,12,18y, 18,26,36,63

3x =y hence,6,8,12,2118,24,36,63

x=6; y=18; x=8; y=24, x=12; y=36; x=21; y=68

3x=yhence, 6,8,12,21,18,24,36,63

3x =yhence, 6,8,12,21,18,24,36,63

x, 6,8,12,18y,18,26,36,63

In ∆ABDAB + BD > AD --1

In ∆ADCAC + CD>AD --2

Adding --1 & --2AB + BD + CA + CD= AD +AD ( BD + CD= BC)AB + BC + CA>2AD

thanks

hit like if you find it useful

Irrational no 1. 1.565665666........2. 1676776777..............Rational no1)3/2

(2x+1)² - (3x+2)² = 0(2x+1-(3x+2)) ( 2x+1+(3x+2))(-x-1) ( 5x+3) = 0sox = -1 and -3/5

The point of intersection is (0,-3)

The equation of the line cutting off positive intercepts in the ratio 4:3

x/4a+y/3b=1................(1)

thus passes through (3,-2)

⇒3/4a+ -2/3a=1multiply both sides by 12a⇒9-8=4a=>a=1/12

4a=4/12=1/33a=3/12=1/4

substituting in eq(1)x/1/3+y/1/4=13x+4y=1or 3x+4y-1=0This is the required equation

The line x + y = a passes through (1, −2)

∴ 1-2=a⇒a=-1

Hence, the equation of the line is x+y=-1

Thanks

2(x-2) -3(x-3) = 5(x-5)2x - 4 - 3x + 9 = 5x - 25-x + 5 = 5x - 256x = 30x = 5

2x-4-3x+9=5x-25-x+5=5x-255x-25=-x+55x+x=25+56x=30x=30/6x=5

Given:

ST || RQ

PS= 3 cm

SR = 4cm

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar(∆PST) /ar(∆PRQ)= (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49

L = length of the straight portion = 90mRadius of the outer perimeter = RRadius of the inner perimeter = r

outer perimeter = 2 L + 2π R = 488 m So R = 49 minner perimeter = 2 L + 2π r = 400 m so r = 35 m

Area of the track (between inner perimeter and outer perimeter)= 2 * L * (R-r) +π (R² - r²)= 6, 216 m²

Cost of development = 6, 216 * 12.50 = Rs 77,700.

Sqroot(5 + sqroot(11+ sqroot(19 + sqroot(29 + sqroot(49))= sqroot(5+sqroot(11+sqroot(19 + sqroot(29+7))= sqroot(5 +sqroot(11+sqroot(19 + 6))= sqroot(5+sqroot(11+ 5))= sqroot(5+4)= sqroot(9) = 3 ans

More number of students lesser number of days.thus they are in inverse proportionrelation of inverse proportion isx1y1 = x2y2100×20=125× y216= y2thus the food provision will last for 16days if 25 more students join the hostel.

volume of pit=LxBxH=3/2×75/100×9=2/3×15/20×9=2/3×3/4×9=1/8×9=81/8m^3; cost=81/8÷450=4556.25

which question do you want??

Post questions one by one

let the principal be XSI isX×2×20/3×100=2X/15C.I isX{(1+20/3×100)²-1}X{(1+1/15)²-1}X{(16/15)²-1}X{256/225-1}X{256-225/225}X{31/225}or 31X/225sum of si & ci is===2X/15+31X/225=48830X+31X/225=48861X/225=488X=488×225/61X=1800hence the pricipal is 1800

3a=4+a2a=4a=23b=a+b+62b=2+6=8b=43d=3+2dd=33c=c+d-12c=3-1=2c=1

THANK YOU

L = length of the straight portion = 90mRadius of the outer perimeter = RRadius of the inner perimeter = r

outer perimeter = 2 L + 2π R = 488 m So R = 49 minner perimeter = 2 L + 2π r = 400 m so r = 35 m

Area of the track (between inner perimeter and outer perimeter)= 2 * L * (R-r) +π (R² - r²)= 6, 216 m²

Cost of development = 6, 216 * 12.50 = Rs 77,700.

first we multiply all number we get 15000 but we have three - sign so we get - 15000

The above multiplication equals 1361.4

the answer is 1361. 4

(1)x^2-x-6=x^2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3),; (2) x^2-4x+3=x^2-3x-x+3=x(x-3)+1(x-3)= (x-3)(x+1); x=-3 & 1

f(x)=x^2-4=(x+2)(x-2); x=-2& 2

sn = n/2 (2a+(n-1)d)

= 12/2 (2(1) + (12-1)*2)

= 6*(2+22)

= 6*24

= 144

Using the quadratic formulax=-b+-√b^2-4ac/2ahencex=5+-√25+40/2x=5+-√65/2hencex=5+√65/2x=5-√65/2

From ΔABC:

tan (θ) = opp/adj

tan (30) = BC/AB

BC = AB tan (30)

From ΔBCD:

tan (θ) = opp/adj

tan (60) = BC/BD

BC = BD tan (60)

Equate the 2 equations:

AB tan (60) = BD tan (30)

Define x:

Let BD = x

AB = x + 20

Solve x:

AB tan (30) = BD tan (60)

(x + 20) tan (30) = x tan (60)

x tan (30) + 20 tan (30) = x tan (60)

x tan (60) - x tan (30) = 20 tan (30)

x ( tan (60) - tan (30) ) = 20 tan (30)

x = 20 tan (30) ÷( tan (60) - tan (60) )

x = 10 m

Find the distance:

Distance = 10 + 20 = 30 m

Find the height:

tan (θ) = opp/adj

tan (60) = BC/10

BC = 10 tan (60) = 10√3 m

120+(3×15)+(19-12)-7×10=120+45+7-70=102

102 is answer this question

sin ϴ + 2 cos ϴ = 1

Squaring both the sides

(sin ϴ + 2 cos ϴ) ² = (1) ²

sin² ϴ + 4 cos² ϴ + 4 sin ϴ cos ϴ = 1

because sin² ϴ = 1 - cos² ϴ & cos² ϴ=1- sin² ϴ

So replacing sin² ϴ by 1 - cos² ϴ and cos² ϴ by 1- sin² ϴ

we get

1 - cos² ϴ + 4 ( 1 - sin² ϴ ) + 4sin ϴ cos ϴ = 1

1 - cos² ϴ + 4 – 4sin² ϴ + 4 sin ϴ cos ϴ = 1

5 – 1 = cos² ϴ +4sin² ϴ - 4 sin ϴ cos ϴ

or

( cos ϴ – 2 sin ϴ ) ² = 4

cos ϴ -2sin ϴ = ± 2 or simply 2 ignoring -2