Explore topic-wise InterviewSolutions in Current Affairs.

i) 6 × 7k = 42k

ii) (-3l) × (-2m) = 6 lm

iii) (-5t²) × (-3t²) = 15t⁴

iv) 6n × 3n = 18n²

thanks

Perimeter of square = 4 × side

a) P = 4 × 10 cm = 40 cm

b) P = 4 × 1.9 cm = 7.6 cm

right

a) 21b-32+7b-20b==21b-32-13b=8b=32, b=32/8=4, (ii)-z^2+13z^2-5z+7z^2-15z= 20z^2-z^2-20z=19z^2-20z; 19z^2=20z;, 19z^2/z=20 ; 19z=20; z=20-19=1;

p(1)= 1-3+2= 2-2= 0

p(0)=0^3-3(0)^2+2(0)

p(0)=0-0+0=0

p (1)=1-3 (1) +2 p (1)=1-3+2 p (1)=3-3p (1)=0

the no is 1007811

12 boys complete in 10 days1 will complete in 10/12 days3 will complete in 10/12 ×3=5/2 days.... (1)8 men complete in 10 days1 will complete in 10/8 days3 will complete in 10/8 ×3=15/4 days..... (2)total days=5/2 +15/4=25/4=6.25 days

Area of shaded region= Area of square-Area of circle=(side)²-πr²=5*5-(3.14)*(2.5)² cm²=25-19.625 cm²=5.375 cm²

a) 500/50=10:1c) 900/27=100:3

(a) 5rs =500 rupees

500 : 50 10:1

(b) 27cm= 270mm

9 : 270 1 : 30

find the ratio of 9 :to 81cm

Let the diameter of Earth be x and its radius x/2.

Let the diameter of moon be 1/4x and its radius be 1/8x.

Volume of earth= 4/3×π×(x/2)³Volume of moon=4/3×π×(x/8)³

Ratio of volumes= {4/3×π×(x/8)³}/{4/3×π×(x/2)³}=1/8

Therefore,1/8 of earth' s volume is moon's volume

THANKS

THANKS SIR

There are9entries, any of which can be0or1. That gives2^9possibilities.

For imagine that you want to produce such a matrix. We will make the first row, then the second, then the third.

For the first entry we have2choices. For every such choice, there are2choices for the second entry. So the first two entries can be chosen in2×2ways. For every choice of the first two entries, there are2choices for the third entry, for a total of(2×2)×2=2^3

Now we start on the second row. For every one of the2^3ways of choosing the first row, there are2choices for the first entry in the second row. So our first four entries can be chosen in2^4ways. Continue.

9 प्रविष्टियां हैं, जिनमें से कोई भी 0or हो सकती है। यह 2 ^ 9 संभावनाएं देती है।

कल्पना करें कि आप इस तरह के मैट्रिक्स का उत्पादन करना चाहते हैं। हम पहली पंक्ति बनाएंगे, फिर दूसरी, फिर तीसरी।

पहली प्रविष्टि के लिए हमारे पास 2 विकल्प हैं। ऐसी हर पसंद के लिए, दूसरी प्रविष्टि के लिए 2 विकल्प हैं। तो पहले दो प्रविष्टियों को 2 × 2 तरीकों से चुना जा सकता है। पहली दो प्रविष्टियों में से प्रत्येक विकल्प के लिए, तीसरी प्रविष्टि के लिए 2 बार हैं, कुल (2 × 2) × 2 = 2 ^ 3 के लिए

अब हम दूसरी पंक्ति पर शुरू करते हैं। पहली पंक्ति को चुनने के लिए 2 ^ 3 में से हर एक के लिए, दूसरी पंक्ति में पहली प्रविष्टि के लिए 2 विकल्प हैं। तो हमारी पहली चार प्रविष्टियों को 2 ^ 4 तरीकों से चुना जा सकता है। जारी रहना।

512 is the answer to that questionOption D

1

2

3

4

Given: In triangle ABC, P and Q are the mid points of sides AB and AC.

To Prove: PQ // (parallel to) BC

Proof: In triangle ABC;AP = PB (coz P is the mid point of AB!)=> AP/PB = 1

Similarly; AQ/QC = 1(Q is the mid point of AC)

Therefore, AP/PB = AQ/QC

​Thus, by the Converse of the Basic Proportionality Theorem,

PQ//BC

Hence Proved.

AB is vertical pole with A at ground. C is midpoint of AB

P is another point in ground and Distance of AP= nAB

Let AB=X

So, AC=X/2

AP= nX

Given that Beta is the angle subtended by BC at P

Assume Alpha is the angle subtended by AC, such that AB subtends an angle of (Alpha +Beta) at P

Therefore, tan (Alpha) = (X/2) / nX = 1/2n

tan (Alpha +Beta) = X/nX = 1/n

We know the formula: tan (x+y) = [tan x + tan y]/ [1- tan x * tan y]

Therefore, 1/n= [tan Beta + 1/2n ] / [ 1- (tan beta*1/2n)]

After solving the above equation,

tan Beta = n/(2n^2+1)

√110=10.4880884817 is the answer

1/(a+b+x) - 1/x = 1/a + 1/b=> -(a+b)/x(a+b+x) = (a+b)/ab. Cancel (a+b) from both sides & cross multiply=> -ab = x(x+a+b)=> x^2+(a+b)x+ab = 0=> (x+a)(x+b) = 0

x = -a or -b Ans

Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(x - π/3) = {1 + cos(2x - 2π/3)}/2

ii) Hence, left side of the given one is:

= {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2

= (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)]

= (3/2) + (1/2)[cos(2x) + 2cos(2x)*cos(2π/3)][Since cos(A+B) + cos(A-B) = 2cosA*cosB]

= (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2]

= 3/2 = Right side

tanx=2tan(x/2)/1-tan^2(x/2)=-4/3; 5 tan^2(x/2)-6tan(x/2)-4=0;(tan(x/2)-2)(2tan(x/2)+1)=0; tan(x/2)=2,-1/2

a)