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The correct answer is1,834

1+3+5+7+9+11+13+15+17+19+21=121

वास्तविक तथा बराबर मूल के लिएD जीरो से बड़ा ,बराबर होना चाहिए

= [ 2(1 + 3k)]² - 4×1×7(3 + 2k) = 0= 4×(1 + 3k)² - 4×7(3 + 2k) = 0= 4(1 + 9k² + 6k) - 4(21 +14k) =0= (1 + 9k² + 6k) - (21 +14k) =0=> 9k² - 8k -20 = 0

now solving the equation we get= 9k² - 18k + 10k - 20 = 0= 9k( k- 2) + 10(k -2) =0=> 9k +10=0 and k -2 = 0=> k = -10/9 and 2 answer

Thanks a lot

(a+b)(a-b)= a² - b². (x²+6)(x²-6)= x⁴ - 36Please hit the like button if this helped you

The product of four consecutive natural numbers is 840 . find the numbers

ram travels 160 kmin 8 hrs what is the speed of his car

distance is 160 kmtime is 8hrsspeed =distance/time=160/8=20km/hr

i) 2/3 = 0.6 bar5/2 = 2.5

so 5/2 > 2/3

5/2 greater becoz value of it is 2.5 and value of 3/2 is 0.6

5 by 22 by 3=0.65 by 2 = 2.5

977 का इकाई अंक = 7 976 का इकाई अंक = 6 999 का इकाई अंक = 9 398 का इकाई अंक = 8 7×6×9×8 = 42×72 = 3024 इकाई अंक = 4

sum of angles of quadrilateral = 360°

=> fourth angle = 360-(72+36+54) = 198°

(2^6×2^30×2^12)/2^32=2^48/2^32=2^16=4^8

64x³ - 125y³ = (4x-5y)³ + 3(4x)(5y)(4x-5y)= (16)³ + 3(4)(5)(12)(16)= 4096 + 11520= 15616.

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Where is the figure?

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thanks

I am trying to do this question for a long time

option ii is the correct answer

Given,ABCD is a quadrilateral in which AB = CD its diagonals AC and BD intersect at O such that OB=OD.

To show:

(i) ar (DOC) = ar (AOB)

(ii) ar (DCB) = ar (ACB)

(iii) DA || CB or ABCD is a parallelogram.Construction,

DE ⊥ AC andBF ⊥ AC aredrawn.

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (each 90°)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

Therefore, ΔDOE ≅ΔBOF

(by AAScongruence rule)

Thus, DE = BF (By CPCT) — (i)

also, ar(ΔDOE) = ar(ΔBOF) ........(ii)

(Two Congruenttriangles have equal areas)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (each 90°)

CD = AB (Given)

DE = BF (From i)

Therefore,ΔDEC ≅ΔBFA

byy RHScongruence rule)

Thus, ar(ΔDEC) = ar(ΔBFA) ........(iii)

(TwoCongruent triangles have equal areas)

Adding (ii) and (iii),

ar(ΔDOE)+ ar(ΔDEC) = ar(ΔBOF)+ ar(ΔBFA)

ar (DOC) = ar (AOB)

(ii)ar(ΔDOC) = ar(ΔAOB)

⇒ar(ΔDOC)+ ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)

(Addingar(ΔOCB) on both sides)

ar(ΔDCB)= ar(ΔACB)

(iii)From part (ii) (ΔDCB) & (ΔACB) have all areas and have the same base BC.So, (ΔDCB) & (ΔACB) must lie between the same parallels.

DA || BC — (iv)

∠FBO= ∠EDO.....(v)

(ΔDOE ≅ ΔBOF )

∠FBA=∠EDC.....(vi)

(ΔDEC ≅ ΔBFA )

On adding eq v & vi

∠ABD=∠CDB

Therfore,DC||AB......(vii)

From eqiv & vii, We get DA||CB & DC||AB

Hence,ABCD is parallelogram .

ii) is the correct answer

why issue give me reason

(a + b)² - ( a -b)²

a² + b² + 2ab - ( a² + b² - 2ab)

a² + b² + 2ab - a² - b² + 2ab

4ab

Thank-you

We know √2 is an irrational number

Let 3 + √2 be a rational number ,

3 + √2 = r , where r is any rational

√2 = r - 3This is an absurd result because LHS is irrational and RHS is rational so this is a contradiction to our assumption, hence3 + √2 is an irrational number

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चार भुजाओं से बने हुए आकृति को चतुर्भुज कहते हैं

Figure having four number of sides is known as tetragon.