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A^2 = A

A^3, A^4, A^5.......A^n = A

(I + A)^n = nC0 I^n + nC1 I^n-1A + nC2 I^n-2 A^2 +...... nCn A^n

= nC0 I^n + nC1 A + nC2 A +...... nCn A

= 1* I + A(nC1 + nC2 +......... nCn)

= I + A[(nC0 + nC1 + nC2 +......... nCn) - nC0]

= I + (2^n - 1)A

1

2

thanks bro

If given terms are in AP

Then,a1 = 2a3 = 2 + 2d26 = 2 + 2d2d = 24d = 24/2 = 12

a2 = 2 + 12 = 14a4 = 2 + 3*12 = 38

Therefore, missing terms are 14 and 38

1350 का गुणन खण्ड करने पर : 1350 = 2×3×3×3×5×5 इसे 2 × 3 से गुना करने पर यह एक पूर्ण वर्ग संख्या बन जाएगी

अतः 6 ही वही संख्या है जिससे 1350को गुना करने पर एक पूर्ण वर्ग संख्या प्राप्त होगी।

2 , 2+12=14 , 14+12=26 18 , 18-5=13 , 13-5=8 , 8-5=3 5 , 5+1.5=6.5 , 6.5+1.5 = 8 , 8+1.5 = 9.5 -4 , -4+2=-2 , -2+2=0 , 0+2=2 , 2+2=4 , 4+2=6 53 , 53-15=38 , 38-15=23 , 23-15=8 , 8-15=-7 , -7-15=-22

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2 metres, 150 cm2 metres = 2 × 100 = 200 cmRatio = 200cm ÷ 150cm = 4 ÷ 3

Ratio = 4 : 3

HCF of fractions =HCF of numerators / LCM of denominators •°•Nume.=HCF of 15 ,12,40 that is 1 •°•Deno.=LCM of 14,35,49 that is 490 so answer is option 2 that is 1/490

1) is the correct answer

I thought correct answer is 2 options

correct answer is 1HCF of( 15 ,12 ,40). 1----------------------------- = ----LCM of (14,35,49). 490

Hope you understood that question

number 2 would be the best choice

f(x) = x^xDF(x)/dx= x^x(log(x)+1)when equals to 0x= 1/e

Too complicated

suppose the angles are x , y and z . z being the greater angle

so, from equation x+y = z now sum of angle = 180° or , (x+y)+z = 180° => z+z =180° => 2z = 180°z = 180/2 = 90°

so, the greatest angle is 90° . hence proved.

tq😇😇

Shaded region = Area of circle - areaof squarenow diameter = diagonal of square

= area of circle = π*2*2- 84.56cm^2

part1

Let share of A=x and share of B=ygiven that2x=3y-200x=3y-200/2:y+3y-200/2=1002y+3y-200=200 (mult by 2 both side)5y=400y=80x=100-80=20

let A= x and B= y; 2x = 3y -200; X= 3y - 200; 2( 3y -200)-3y = -200;; 6y - 400 - 3y = -200;; 3y = -200 +400;; y=200/3=66.66

Let A’s share be x.

Then B’s share will be (100 - x)

According to the question -:

Twice of A’s share, that is 2x

Is less than, that is (- 2x)

3 times B’s share that is 3(100 - x)

Is 200.

Now, let us write the equation as per the question. We know all the terms, right? Let us put them together.

3(100 - x) - 2x = 200

Now we know the equation, let us solve it. On solving the bracket, we will get - :

300 - 3x - 2x = 200

300 - 5x = 200

-5x = 200 - 300

-5x = - 100

Now, (-) and (-) gets cancelled, we get - :

x = 20

That is A’s share was 20 and that of B’s was 80

tan370*cot10=tan370/tan10

tan370=tan(360+10)=tan10

tan370/tan10=tan10/tan10 =1

370-0=370 (basic subtraction)

x=37y=53

x = 37y =53is the correct one

Between 2 and 3 there are infinite rational numbers as we can write decimals numbers between 2 and 3 in form of p/qsuch that q is not 0

Rational number between 3/7 and 2/3 = (3/7 +2/3)/2 = (9+14)/42=23/42

(3/16) +(-3/16) 0 answer

0 is the correct answer of this question

0 is the correct answer of the given question

0is the answer of the question

As the unit ofvolume given in the question is too small so considering it 6160 m³Volume of cylindrical tank =π r²×h⇒6160=π×(14)²×h⇒h = 10 mnow, the area of curved surface= 2π r×h =2π×14×10 =880 m²cost of painting that surface = Rs 3×880 = Rs2640