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L.H.S= (m^2+n^2)cos^2b

= {(cosa/cosb)^2+(cosa/sinb)^2}cos^2b

= {(cos^2a/cos^2b) +(cos^2a/sin^2b)}cos^2b

= (cos^2a.sin^2b +cos^2a.cos^2b / cos^2b.sin^2b)cos^2b

= cos^2a(sin^2b+cos^2b) / sin^2b

= cos^2a / sin^2b

R.H.S= n^2

=(cosa/sinb)^2

= cos^2a / sin^2b

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48622530 is the right answer....

Total sale receipt in two years is = Rs 20956480 + Rs 27666050 = Rs 48622530

total 1500 B + G = 1500 G = B + 50 B + B + 50 = 1500 2B = 1450 B = 725 G = 775

For first year = 100000(1+(5/100)) =100000*21/20 =105000

For second year = 105000(1+(8/100)) = 105000*27/25 = 113400

The Cost of land after 2 years will be Rs.113400

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If one of the zero is reciprocal of the other,let one zero beαother zero will be 1/α

Product of roots = c/a⇒α× 1/α = 6a/(a²+9)⇒ 1 = 6a/(a²+9)⇒ a² + 9 = 6a⇒ a² - 6a + 9 = 0⇒ a² - 3a - 3a + 9= 0⇒ a(a-3) -3(a-3) = 0⇒ (a-3)(a-3) = 0⇒ (a-3)²= 0⇒ a = 3

Value of a is 3.

1 km = 1000 m 1 hr = 3600 seconds1) 118 km/h = 118×1000/3600 = 295/9 m/s 2) 5.15 km/h = 5.15×1000/3600 = 51.5/36 m/s

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the divident should be 4z^3 + 14z^2 + 6z

ghj CD r be ebjdbf fbjdnddhjd d DVD d xhjx d Dhba d find d his d bcjx X xjks d hi PC dvxjd d xhd km cbcjdn f by ksnxjbxj c cjc f

we should __________ in the library

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Let cosec-¹(2) = ∅ where ∅ is the principal branch of cosec-¹(2) therefore cosec∅ = 2 since the range of principal value of branch of cosec-¹x is [+π/2,π/2] -{0} therefore -π/2≤∅≤π/2, ∅≠0

Now cosec∅ = 2 = cosecπ/6 thus ∅ = π/6 as cosecπ/6 = 2 and π/6 € [-π/2,π/2]-{0}

we have to find at which time all will ring together so we have to find L.C.M. of all 4 numbers8=2×2×29=3×312=3×2×215=3×5so L.C.M.=2×2×2×3×3×5=360so after 360 seconds they all toll.

18 seconds will they toll together

They both need to be toll together, so we have to take LCM of both the timingsso here that is 360sso after 3 minutes both will toll together.so at 12:03 they both will toll together.

Aprimary key, also called aprimarykeyword, is akeyin a relational database that is unique for each record. It is a unique identifier, such as a driver license number, telephone number (including area code), or vehicle identification number (VIN). A relational database must always have one and only oneprimary key.

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(1/3)x=(2/7)x+1(1/3-2/7)x=1(7-6)x/21=1x/21=1x=21Explanation:1/3 of 27 is 7, which exceeds 2/7 of 27 (that is 6) by one

Solution :-

First of all we have to find the square root of 2509 using the long division method.

50 ____________ 5 | 2509 5 | 25 |______100| 09 |

Here, the remainder is 9.

Hence, 50² < 2509

And we know that, 50² = 2500

Hence, the number to be subtracted is 2509 - 2500 = 9

So, 9 should be subtracted from 2509 to make it a perfect square.

after subtracting 9 in 2509 it become perfect square of 50

(1/3)x=(2/7)x+1(1/3-2/7)x=1(7-6)x/21=1x/21=1x=21Explanation:1/3 of 27 is 7, which exceeds 2/7 of 27 (that is 6) by one

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749 hai iska answer to

1875/16200=(75×25)/(648×25)=(25×3)/(216×3)=25/216(basic multiplication)