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=ab-ac+bc-ab+ac-bc=0

=ab-ac+bc-ab+ac-bc=0

ab-ac+bc-ab+ac-bc=ab-ab+bc-bc+ac-ac=0

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can u solve this

answer is - a five hundred fifty

c (x+y) + d (x+y)=(x+y)(c+d) it's answer like it hope this helps u

(x+y)(c+d) is right answer

(x+y((c+d) is right answer

12cd+12df this is right answer of this question

use (a+b)²&(a+b)²=c²+(3d+f)² + 2c(3d+f) - [c²+(3d+f)² - 2c(3d+f) ]

=c²+(3d+f)²+ 2c(3d+f) - c² - (3d+f)² +2c(3d+f)

=4c(3d+f) answer

sum of n numbers = n(n+1)/2therefore sum of 40 no.s= 40*41/2= 820sum of 15 no.s = 15*16/2 = 120

16+17+18+....+40= sum of 40 no.s - sum of 15 no.s = 820-120=700.

The right answer.i.e., 700 is not mentioned among the options, but it is the right answer.

852=2×2×3×711491=3×7×71HCF=3×71=213LCM=2×2×3×7×71=5964LCM×HCF=1270332Product of two number=852×1491=1270332

Using, sin(C) + sin(D) = 2[sin{(C + D)/2}*cos{(C-D)/2}],

sin(6x) + sin(2x) = 2sin(4x)*cos(2x)

2) HENCE, sin(2x)+sin(4x)+sin(6x)=0, ==> 2sin(4x)*cos(2x) + sin(4x) = 0

3) ==> sin(4x){1 + 2cos(2x)} = 0

4) ==> Either sin(4x) = 0 or 1 + 2cos(2x) = 0

5) When sin(4x) = 0, the general solution for 4x = 2kπ, where 'k' is an integerSo, 'x' = kπ/2

6) When 1 + 2cos(2x) = 0, cos(2x) = -1/2

==> the general solution for 'x' = (2k + 1)(π/2) ± (π/6)

a^2b^2x^2 + b^2x - a^2x - 1 = 0

b^2x(a^2x - 1) - 1(a^2x + 1) = 0

(b^2x - 1)(a^2x + 1) = 0

x = 1/b^2, - 1/a^2

Perimeter of garden = 2×(L+B)= 2× (120 + 100) = 2× 220 = 440 m

To cover 2200 m boy needs = 2200m/440m= 5 So, 5 rounds needed

Perimeter=2(l+b)160=2(l + 10)160=2l+20160-20=2l140=2l140/2=l70m

Area of garden= l×b= 70 ×10= 700m2

the answer should be 1

2(cos²∅-sin²∅) = 1

=> 2cos2∅ = 1..... ( since cos²∅-sin²∅ = cos2∅)

=> cos2∅ = 1/2

so. 2∅ = 2nπ + π/3

=> ∅ = nπ + π/6

Given, Median of Nine observations = 20

Median of 9 observations = 9+1/2 th term = 5th term.

Now, Fifth term = 20

The last term or the largest four are increased by 2 .But the fifth term is unchanged.

The median of resultant set is same as the earlier.

Hence median of resultant set = 20

The Median remains Unchanged. The number of Observations is odd, hence the fifth observation will be the median.hence , no effect of altering largest 4 observations.

Solution no. 1:-Arithmetic Mean = ∑X/NGiven : Mean = 50 and N = 100Therefore,50 =∑X/100∑X(Wrong) = 5000Correct value = 110Incorrect value = 100Correct Arithmetic Mean = {∑X (Wrong) + Correct valueof observation- Incorrect value of observation}/N⇒ Correct Mean = (5000 + 110 - 100)/100⇒ Correct mean = 5010/100⇒ Correct Mean = 50.1AnswerSolution no. 2 :-In this question, 'the value of thelargest item' should be written instead of 'latest item'.The value of the Median will not change because whatever values are added or subtracted from median, the total observation will remain 100 and median is the centrally located value of a series such that the half of the value or items of the series are above it and the otherhalf arebelow it.Formula of median = Size of(N+1)/2th ItemTotal observations = 100Size of (N+1)/2th item⇒(100+1)/2th Item⇒ 101/2⇒ 50.5th Item110 is corrected observation instead of 100 and 110 is the largest of all the observation. So it will not make any difference in value of median. The value of Median will be 52.Answer

Correct mean = (50*100 + 110-100)/100= 50.1

Median would not be varying due to the changed observation. So, median = 52.

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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 HENCE PROVED

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sin^6x + cos^6x

(sin²x)³ + (cos²x)³

Use a³ + b³ = ( a + b) ( a² + b² - ab)

(sin²x + cos²x) ( sin⁴x + cos⁴x - sin²cos²x)

((sin²x+cos²x)² -2 sin²x cos²x - sin²cos²x)

( 1 - 3 sin²x cos²x)

Proved

2(1-sin^x) + 3sinx = 0-2sin^2x + 3sinx + 2 = 0

2sin^2x -3sinx - 2 = 0(2sinx + 1)(sinx - 2) = 0sinx = -1/2 or sinx =2 ---but sinx cannot = 2 so just work with -1/2

sinx = -1/2 quadrants III and IVx = 210 deg (or 7pi/6) + 2npi in QIIIx = 330 deg (or 11pi/6) + 2npi in QIV

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this result is from Demovier Theoremhence as n=32cos3theta

l not understand

Given, dimensions of rectangular park = 50mx40mSo, Are of park = 50*40 = 2000sq.mArea of the grass strip surrounding the pond = 1184 sq.mHence, area of the pond = lb = 2000-1184 = 816sq.m ⇒ lb = 816sq.mLet the width of the grass strip be 'x'So, (50-2x)(40-2x) = 816⇒ 2000 - 100x - 80x + 4x² = 816⇒ 2000-180x+4x² = 816⇒4x²-180x+1184 = 0⇒x²-45x+296 = 0⇒x²-37x-8x+296=0⇒x=37, x=8But x=37 is not possible∴ Width of the grass strip = 8mLength of the pond = 50-16 = 34mBreadth of the pond = 40-16 = 24m

n/5 -5/7 = 2/3

n/5 = 2/3 + 5/7

n/5 = (14 + 15)/21

n/5 = 29/21

n = (29×5)/21

n = 145/21

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Given the ratio of length : width = 5:2 so in fraction it can be written as length/width = 5/2 but width is given as 40m. so length/40 = 5/2 => length =( 5/2)*(40) THEREFORE , length = 100m.

2/7 + 6/7(2+6)/7 = 8/7

integration of [ e^x { f(x) + f'(x) } ] = e^x * f(x)If f(x) = sec x, then f'(x) = sec x * tan x

integration of [ e^x * sec x * (tan x + 1) ] = integration of [ e^x ( sec x + sec x * tan x ) ]= integration of [ e^x { f(x) + f'(x) } ]= e^x * f(x) + C= e^x * sec x + C

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