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X^2+4x+4= (x+2)^2now(x+2)^2/(x+2)= (x+2)thanks

option d is correct5/100

Given:In the given figure O is the centre of the circle.To Prove:∠AOC=∠AFC+∠AECProof:In ΔBEC using exterior angle theorem.Exterior angle theorem property:Sum of two interior angle of triangle is equal to opposite exterior angle. So, we get∠ABC=∠AEC+∠BCD Double the above equation both so,2∠ABC=2∠AEC+2∠BCDAngle subtended on circle is half angle subtended at centre. 2∠ABC=∠AOC∠AOC=∠AEC+∠BCD+∠AEC+∠BCD∠AOC=∠AEC+∠BCD+∠ABCBut∠ABC=∠ADC(∴Angles subtended on same arc are equal)∠AOC=∠AEC+∠BCD+∠ADCIn ΔFDC using exterior angle theorem.∠AFC=∠BCD+∠ADCHenceproved,∠AOC=∠AFC+∠AEC

When ,(5m+1)(5m+3(5m+4) 5=5m(5m+3)(5m+4) + 1(5m+3)(5m+4) 5

=5m(5m+3)(5m+4) + 5m(5m+4) + 3(5m+4) 5={ m(5m+3)(5m+4) + m(5m+4) +15m + 12}/5 = {m(5m+3)(5m+4) + m(5m+4) + 3m + 12}/5

The remainder of 12/5 is 2, so the remainder of (5m+1)(5m+3)(5m+4)/5 is 2.

in the figure <APC is equal to <BPDP bisect AB so AP=BP....similarly CP=DP......in triangle APC and triangle BPD ..AP=BP (given). CD=DP(given ).and angle BPD =angle APC (vertically opposite angle) .so by SAS criteria triangles are congruent

In ∆APC and ∆DPBangle APC = angle DPBAP = PDPC = PDby SAS rule ∆APC is congurent to ∆BPD

By angle bisector theoremMN:NP=MQ:QPTherefore,MQ:QP=4:11And hence,MN:NP=4:11

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o , -3 is the distance from x

[(12^1/5)/(27^1/5)]^5/2

= (12/27)^[1/5*5/2]

= (12/27)^1/2

= (4/9)^1/2

= 2/3

perpendicular distance will be equal to X component that is 5hence distance is 5 units.

Explanation:Perpendicular distance of the point P(x,y) from the y-axis = ordinate of the point P(x,y) = x.

how sir

The perpendicular distance is the y coordinate.So the distance is 12.

nice

please tell me

=>(1/2+1-2/√3)/(2/√3+1/2+1)

=>(3/2-2/√3)/(3/2+2/√3)

=>(3√3-4)/(3√3+4)

=>(3√3-4)^2/(27-16)

=>(27+16-24√3)/11

=>(43-24√3)/11

Thankyou

p =2

4 units because the perpendicular distance is the x Co-ordinate of that point

What is answer??????

1 /30 is thw right answer

one more question I doubt

a) is the correct answer

a) is the right answer of the following

a) is the right answer of the following

option A is the correct answer

Take LCM(x^3-8)/(x-2).......(1)(x^3-2^3)/(x-2)a^3-b^3=(a-b)(a^2+b^2+ab)x^3-2^3=(x-2)(x^2+4+2x)equation (1) becomes(x-2)(x^2+4+2x)/(x-2)(x-2) gets cancelled=x^2+4+2xx=-1+√3ix=-1-√3i

equation of y axis is x= 0so it will be same as y coordinate so it will be 13 unit

4.5

Sin⁴x/a + cos⁴x/b = 1/(a + b)

we know, sin²x + cos²x = 1 ⇒cos²x = 1 - sin²xcos⁴x = (1 - sin²x)² = 1 + sin⁴x - 2sin²x , use it above

sin⁴x/a + (1 + sin⁴x - 2sin²x)/b = 1/(a + b) sin⁴x/a + 1/b + sin⁴x/b - 2sin²x/b = 1/(a + b) sin⁴x(1/a + 1/b) -2sin²x/b = 1/(a + b) - 1/b sin⁴x(a + b)/ab - 2asin²x/ab = (b - a - b )/(a + b)b (a + b)²(sin²x)² - 2a(a + b) sin²x = -a²{(a + b)sin²x}² -2.a.(a + b)sin²x + a² = 0 [ this is like (a - b)² = a² - 2ab + b² ]{(a + b)sin²x - a}² = 0sin²x = a/(a + b) ⇒1 - sin²x = cos²x = 1 - a/(a + b) = b/(a + b) hence, sin²x = a/(a + b) and cos²x = b/(a + b)

so, sin⁸x = a⁴/(a + b)⁴ and cos⁸x = b⁴/(a + b)⁴

now, sin⁸x/a³ = a/(a + b)⁴ cos⁸x/b³ = b/(a + b)⁴

So, sin⁸x/a³ + cos⁸x/b³ = a/(a + b)⁴ + b/(a + b)⁴ = (a + b)/(a + b)⁴ = 1/(a + b)³

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In this question we have to find cos²A in terms of m and n , so we have to eliminate ∠B from the given relations.

tan A = n tan Btan B = 1/n tan A

Cot B = n /tan A [ cot B = 1/tan B]

sin A = m sinB sin B = 1/m sinAcosec B = m / sinA [sinB = 1/cosecB]

cosec²A - cot²B =1

Substitute the value of cot B and cosec B in the above relation.

(m / sinA)² - (n /tan A)²(m² / sin²A) - (n² /tan² A)(m² / sin²A) - (n² /(sin²A / cos²A))

[ tan A = sinA / cosA]

(m² / sin²A) - n²cos²A / sin²A = 1m² - n²cos²A = sin²Am² - n²cos²A = 1- cos²A

[sin²A = 1- cos²A]

m² -1 = n²cos²A - cos²Am² - 1 = cos ²A(n² -1)

cos²A = m² -1/ n²-1

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in two triangles ABC and abdap=pb. as p bisects absimilarly cp=pdangle apc=bpd (vertically opp. angles) therefore triangles are congruent (SAS RULE

in these two triangles angle APO will be equal to angle DPBas both are vertically opposite triangleAP= PBPC= PDso with the help of SAS these two triangles are congruent

it is triangle apc and bpd

2x/a+y/b+x/a-y/b=2-4=-23x/a=-2so x=-2a/3-2a/3a-y/b=4so y/b=-2/3-4=-14/3

first information: 100-80=20 females

2nd information:out of 60= 30 males hence = 60-30=30 females

hence the given information is not correct

sorry this Q. is from triangles chapter

The answer to this question is there exists only one point. And the co-ordinates of that point are (x1+X2+X3),(y1+y2+y3)

cot (90-A) = cos(90-A)/sin(90-A)x sin (90-A)* cot(90-A) = x sin(90-A)*[cos(90-A)/sin(90-A)]= x cos(90-A)= cos(90-A) when x = 1.

1

1

Given,x = a cos^3 thetax/a = cos^3 theta.... (1)

y = b sin^3 thetay/b = sin^3 theta.... (2)

Then, (x/a)^2/3 + (y/b)^2/3= (cos^3 theta)^2/3 + (sin^3 theta)^2/3= cos^2 theta + sin^2 theta= 1

Given:In the given figure O is the centre of the circle.To Prove:∠AOC=∠AFC+∠AECProof:In ΔBEC using exterior angle theorem.Exterior angle theorem property:Sum of two interior angle of triangle is equal to opposite exterior angle. So, we get∠ABC=∠AEC+∠BCD Double the above equation both sides2∠ABC=2∠AEC+2∠BCDAngle subtended on circle is half angle subtended at centre. 2∠ABC=∠AOC∠AOC=∠AEC+∠BCD+∠AEC+∠BCD∠AOC=∠AEC+∠BCD+∠ABCBut∠ABC=∠ADC(∴Angles subtended on same arc are equal)∠AOC=∠AEC+∠BCD+∠ADCIn ΔFDC using exterior angle theorem.∠AFC=∠BCD+∠ADC∴∠AOC=∠AEC+∠AFCHenceproved,∠AOC=∠AFC+∠AEC

hi ju

6x^ 2 - 20x is the correct answer of the given question