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Amount of sugar purchased= 200kgC.P. of 1kg= Rs. 15So, CP of 200kg= 200*15= Rs. 3000Now profit%= 5%Profit%= Profit×100/CP5= P*100/3000 [Substituting values]P*100= 5*3000P= 5*3000/100P= Rs. 150Now, Profit= SP-CPTherefore, SP= Profit+CPSP= 150+ 3000SP= Rs. 3150 for 200 kgSP for 1 kg=3150/200=15.75 Ans

15*5=7575÷100=0.75total prize 15 +.075=15.75ans- C

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sin80° = cos10tan 10°= cot80tan 45° = 1tan40° = cot 50° and sec^theta- tan^2 theta= 1

A+B+C=4x³+3x-19+x³+8x²-7x+13-5x³-6x+20=8x²-10x+14A+B-C=4x³+3x-19+x³+8x²-7x+13+5x³+6x-20=10x³+8x²+2x-26A-B-C=4x³+3x-19-x³-8x²+7x-13+5x³+6x-20=x³-8x²+16x-42

A. ab(4b+3a)B. 8xy -2xy = 6xyC. (5+1)x³ = 6x³D. -xy(2x+5)

Cost price of 1 kg potato, CP= Rs 4

gain= 10% of CP = Rs 0.4

Selling price per kg = CP + gain= Rs 4.4

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Tan (a + b) = 1 => tan (a - b) = 1/7

tan 2 a = tan (a+b + a -b) = [ tan(a+b) + tan (a-b) ] / [1 - tan (a+b) tan (a-b) ] = (1 + 1/7) / (1 - 1/7) = 4/3tan 2a = 2 tan a / (1 - tan² a) = 4/3

=> 2 tan² a + 3 tan a - 2 = 0=> tan a = [-3 + - √[ (9 + 16) ] / 4 = -2 or 1/2

Now we have tan (a+b) = [tan a + tan b] / [ 1 - tan a tan b ]If tan a = -2, then => 1 = [ -2 + tan b ] / [ 1 + 2 tan b ] => tan b = -3

if tan a = 1/2 then, => 1 = [ 1 /2 + tan b ] / [ 1 - 1/2 tan b ] => tan b = 1/3

Tan (a + b) = 1 => tan (a - b) = 1/7

tan 2 a = tan (a+b + a -b) = [ tan(a+b) + tan (a-b) ] / [1 - tan (a+b) tan (a-b) ] = (1 + 1/7) / (1 - 1/7) = 4/3tan 2a = 2 tan a / (1 - tan² a) = 4/3

=> 2 tan² a + 3 tan a - 2 = 0=> tan a = [-3 + - √[ (9 + 16) ] / 4 = -2 or 1/2

Now we have tan (a+b) = [tan a + tan b] / [ 1 - tan a tan b ]If tan a = -2, then => 1 = [ -2 + tan b ] / [ 1 + 2 tan b ] => tan b = -3

if tan a = 1/2 then, => 1 = [ 1 /2 + tan b ] / [ 1 - 1/2 tan b ] => tan b = 1/3

3,6,9,12,15,18,21,24,27

2,6,10,14,18,22,26,30,34

2,4,8,16,32,64,128,256,512,1024

7,11,13,16,20,22,25,29,31,34,37

(X )option b is the correct answer of the given question.

option (B) is the correct answer

1-The gram panchayat is divided into wards and each ward is represented by a Ward Member or Commissioner, also referred to as a Panch or Panchayat Member, who is directly elected by the villagers. 2-The panchayat is chaired by the president of the village, known as a Sarpanch.

Suppose two rupee coin is xthen 1 ruppe coins= 3x5 rupee coin = 2xaccording to question1.(3x)+5(2x)+2(x)= 6003x+10x+2x= 60015x= 600x= 40so number of 1 rupee coin= (3*40)= 1205 rupee coins( 2*40=80)2rupee coins=(40)

8 kg of sugar = 260 => 1 kg of sugar = 260/8 = 32.5

Now 10kg of sugar = 10×32.5 =325

10kg sugar cost 325 rs

8kg of sugar = 260 so 1 kg of sugar = 260/8 = 32.5

now 10 kg of sugar. = 10× price of 1 kg of sugar

=> price of 10 kg = 10×32.5 =325rupess.

Let the number of coins be x.

2 rupee coin be 2x.

5 rupee coin be 5x.

1 rupee coin with the total number of coins is 40 will be 40 - 2x.

Given value is 120.

On solving (1), (2),(3),(4), we get

2x + 5x + (40-2x) = 120

2x + 5x + 40 - 2x = 120

5x + 40 = 120

x = 80/5

= 16.

The number of 2 coins and the 5 rupees coins will be x = 16.

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0.4 is greater than 0.30.07 is greater than 0.023 is greater than 0.80.5 is greater than 0.05

So option c is correct

A+B+C = 180 (ASP of trianlge)

~ B + C = 180 - A

~ cos (B + C) / 2 = cos (180 - A) / 2

~ cos (180/2 - A/2)~ cos (90 - A/2)

~ sin (A/2)

Therefore, cos (B + C/2) = sin (A/2)

b+c = 90-atherefore, its tan (45-(a/2))that is (1-tan(a/2))/(1+ tan(a/2))

1kg is 1000gm1gm is 1000mg

1kg=1000gm1gm=1000mg

Question you have submitted is incomplete. Please post a complete question.

1Gram atom means 1 mole

No. Of moles = Weight in gram/Molar mass

No. Of moles =52/32

=1.625moles

1.625 gram atom (Moles)of sulphur is there in 8 gm of sulphur

In order to compare fractions denominator should be equal

(i) 1/2 of kilogram = 6/12 of kilogramTherefore 6/12 or 1/2 of kilogram is more than 4/12 of kilogram

(ii) 3/4 of metre = 12/16 of metreTherefore 12/16 or 3/4 of meter is more than 6/16 of metre.

(iii) 2/5 of article = 4/10 of article Therefore 6/10 of article is more than 4/10 or 2/5 of article

1

1

2

1 cubic centimetre=

1 × 10-6cubic metre= volume= 1/8*10^-6m^3

AM of number a and b = (a + b)/2 GM of number a and b = √ab

Here, AM : GM = m : n (a + b)/2 : √ab = m : n (a + b)/2√ab = m/n Applying componendo and dividendo rule,(a + b + 2√ab)/(a + b - 2√ab) = (m + n)/(m - n)(√a² + √b² + 2√ab)/(√a² + √b² -2√ab) = (m + n)/(m - n) (√a + √b)²/(√a - √b)² = (m + n)/(m - n) take square root both sides,( √a + √b )/(√a - √b) = √(m + n)/√(m - n) again applying componendo and dividendo,( √a + √b + √a - √b)/(√a + √b - √a + √b) = {√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }2√a/2√b = {√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }√a/√b = {√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }taking square both sides, a/b =[{√(m + n) + √(m - n)}/{√(m + n) - √(m - n) }]²a/b = {m + n + m - n - 2√(m² - n²)}/{m + n + m - n -2√(m² - n²)}a/b = {2m + 2√(m² - n²)}/{2m - 2√(m² - n²)}a/b = {m + √(m² - n²)}/{m - √(m² - n²)}

hence, a : b = m + √(m² - n²) : m - √(m² - n²)

GIVENif a+8√5b=8+√5/8-√5+8-√5/8+√5RHS= 8 + (√5 / 8) - √5 + 8 - (√5 / 8) + √5MAKE IT EQUAL TO a + 8√5b8 + (√5 / 8) - √5 + 8 - (√5 / 8) + √5 = a + 8√5bLCM is 8(64 + √5 - 8√5 + 64 - √5 + 8√5) / 8 = a + 8√5b128 / 8 = a + 8√5b16 = a + 8√5bCan be written as16 + (8√5) x0 = a + 8√5bTherefore, a = 16 & b = 0

i) x^10/10

ii) x^1/3+1/(1/3+1)

= 3x^4/3 / 4

iii) x

yes, this a correct question

yes , this is a correct answer

mane money, time, Dr

yes, this is a correct answer

there are infinitely many such a solution of equation

perimeter = 2*(5x+4) +2*(x-4) = 72=> 10x+8+2x-8 = 72=> 12x = 72=> x = 72/12 = 6

so, length = (5*6+4) = (30+4) = 34m breadth = (6-2) = 2m

x/6-3= 12x-18= 36x= 36+18= 54