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a=6d=9-6=3s9 28th term=a+27d=6+27*3=6+81=87

answer too 21 h

6=3*218=6*372=18*4x=72*5=360

Here the terms are in G.P

I.e.,2,6,18,54

a1=2a2=6a3=18a4=54

a2/a1= 6/2=3

a3/a2=18/6=3

a4/a3=54/18=3

Since ratio is equal, so this is in G.P.

To find the nth term in G.P tn = ar^(n-1)

Here a=2, r=common ratio=3n = 9

By substituting in Formula we get, 2(3)^(9-1)= 2(3)^8= 2(6561)= 13122

9th term for given series is 13122

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[(-2/3)^-3 - (-3/4)^-3] ÷ (3/2)^2

= (-3/2)^3 - (-4/3)^3] ÷ 9/4

= [(-27/8) - (-64/27)] ÷ 9/4

= [(-27*27) - (-64*8)]/8*27 ÷ 9/4

= (729 + 512)/8*27 ÷ 9/4

= 1241/8*27 * 4/9

= 1241/2*27*9

= 1241/486

= 4.33

P(a,b) , in fourth quadrant the y coordinate is negitive and x axis is positive thereforea greater than b

(8a^3 ÷ 27x^-3)^2/3 * (64a^3 ÷ 27x^-3)^-2/3

= (2a * 3x)^[3*2/3] * (4a * 3x)^[3*-2/3]

= (6ax)^2 * (12ax)^-2

= 6a^2x^2/ 12a^2x^2

= 1/2

Let the angles be X and X+10. So, 77+57+X+X+10=360=> 2x = 360-10-57-77=> 2x = 216. => X = 108°X+10 = 118°

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(cosx/2 + sinx/2)^2 =1+sinx

5 = √(4+1)Here 4 and 1 both are perfect square number as √4 = 2 and √1 = 1So draw right triangle with side 2 and 1And according to Pythagoras theorem hypotenuse will √(22+12) = √5Then draw an arc of √5 on number line

let number be xone fourth of a number is added to itself be ( x + x/4)result is 75: x + x/4 = 75 take lcm in left hand side( 4x + x)/ 4 = 755x / 4 = 755x = 75 ×45x = 300x = 300/5x = 60the number is 60

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(3√3+7√2)+(√3-5√2)=3√3+7√2+√3-5√2=3√3+√3+7√2-5√2=√3(3+1)+√2(7-5)=4√3+2√2 (Ans)

4root3 + 2 root2 is the correct answer for this question

in 4th quadrantX coordinate is positive y is negativehence a>b

Thanks

Proof,In ΔABC & ΔABD,AB = AB (Common)AC = AD (Given)∠CAB = ∠DAB (AB is bisector)

Hence, ΔABC ≅ ΔABD. (by SAS congruence rule)

Then, BC= BD (by CPCT)

Thus, BC & BAD are equal.

Number of even numbers on die = 3

the answer is 3/61/2

thanks

(7 + 3root(5)) (7 - 3root(5))

Using(a + b)(a - b) = a^2 - b^2

= (7)^2 - (3root(5))^2

= 49 - 9*5

= 49 - 45

= 4

69 995 10 Answer of a question

6999510 answer of a question

69956. ans.

1. 5i(-3/5)i = -3i² = 3

4. 3(7+7i) +i(7+7i) = 21+21i+7i-7 = 14+28i

6. (1/5+2i/5) - (4+5i/2) = (1/5-4)+(2i/5-5i/2)= -19/5-21i/10

8. (1-i)⁴ = ((1-i)²)² = (1+i²-2i)² = (-2i)² = -4

Product of two numbers = One number * OTher number=> Other number = Product / One number= 1785483/987= 1809.

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Slopes ofx−y=4

=>tanθ=1

=>(sinθ=1/√2,cosθ=1√2)or(sinθ=−1/√2,cosθ=1/√2)(sin⁡θ=−12,cos⁡θ=12)

Q is(2+2√3–√(1/√2),1+2√3(-1/√2))

=(2−√6,1-√6)

equation of required line isx+y=3−2√6

angle y= 125°+141°=266°(since AB is parallel to EF)angle x= 360°-angle y=360-266°=94°

HAl

square of 13 is 3.6055

169 is sq cm of root 13

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Thank you so much

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cotƟ – tanƟ =2cos2Ɵ – 1/sinƟcosƟ

LHS=cosƟ/sinƟ – sinƟ/cosƟ

=cos2Ɵ – sin2Ɵ/sinƟcosƟ

=cos2Ɵ – (1 – cos2Ɵ)/sinƟcosƟ

=cos2Ɵ – 1 + cosƟ/sinƟcosƟ

=2cos2Ɵ – 1/sinƟcosƟ

LHS:- tan theta + sec theta - (sec²theta - tan² theta /tan theta - sec theta + 1 =( tan theta + sec theta)(1- sec theta + tan theta )/tan theta +1 - sec theta =tan theta +sec theta =sin theta / cos theta + 1/cos theta = 1+ sin theta / cos theta ⁼sec theta+ tan thetaLHS = RHS Hence proved

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Part 2 by doing rationalization we proved it Please like the solution 👍 ✔️