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cos 4x hhhhhhhhjjjhj

The first step is to factorise both numerator and denominator.

Both are quadratics and can be factorised using the a-c method.

⇒x2−3x+2=(x−2)(x−1)

andx2+5x−6=(x+6)(x−1)after dividing answer will be (x-2)/(x+6)

Consider x⁴+ x²+ 1 = (x⁴+ 2x²+ 1) – x² = [(x²)²+ 2x²+ 1] – x² = [x²+ 1]²– x²

It is in the form of (a²– b²)

= (a + b)(a – b)Hence [x²+ 1]²– x²= [x²+ 1 + x] [x²+ 1 – x]

what we do in this question

x²-7x-2x+14=0x(x-7)-2(x-7) =0(x-2)(x-7) =0x=2 , x= 7

volume of 6 cm cube= 6*6*6= 216volume of 2 cm cube= 2*2*2=8so no of 2cm cubes from 6cm side cube= volume of 6cm cube/ volume of 2cm cube= 216/8 = 27 cubes

thnx

value of other side is 6.6 cm

because 4.4*3/2=6.6

perimeter = 2(l+b)= 2(4.4+6.6)=22 cm

right ans is 22 cm

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hello this is the answer which I am posting site answer is 22cm

we know that :

First side= 48

Other side= 4.8 * 3/2 = 7.2

Perimeter = 2( 4.8 + 7.2)

Perimeter = 24

the perimeter = 22cm is right answer

permiter of parallelogram = 22cm

Solution:

Given: In quadrilateral ABCD, AB smallest & CD is longest sides.

To Prove: ∠A>∠C & ∠B>∠D

Construction: Join AC. Mark the angles as shown in the figure..

Proof:In △ABC , AB is the shortest side.

BC > AB ∠2>∠4 …(i) [Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD ∠1>∠3 …(ii) [Angle opposite to longer side is greater]

Adding (i) and (ii), we have

∠2+∠1>∠4+∠3 ⇒∠A>∠C

Similarly, by joining BD, we can prove that ∠B>∠D

No. of students in one row = xtherefore no. of rows in the hall = x

total no. of students = no. of students in one row × no. of rows= x × x = x²therefore 8649 = x²x = √8649 = 93

therefore 93 students in each row of the hall

2 and 3 represent the same rational number.

ṭһıś

8x³ + √64 y³

8x³ + 8y³

8( x ³ + y³)

8 ( x + y) ( x² + y² - xy)

angle DRE = 180 - (70+40) = 70angle ARE = 180- 70 ( linear pair) = 110

x²+1/x²=6sq both sides,(x²+1/x²)² = (6)²(x²)²+2×(x²)×(1/x²)+(1/x²)²=36then we cut x² and 1/x²x⁴+1/x⁴=36-2x⁴+1/x⁴=34

No i have problem see Q properly

Sum of remaining angles=360°-55°=305°

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monomial =2xbinomial=3a^2b-2ab^2trinomial=3x^2-2x-5

4 cards in each stakehence in 29 stakes=29*4=116 cards.

Thanks!❤

Total student=48Above avg=1/6 of 48=8 studentsAverage=1/4 of 48 = 12 studentsSo below avg = 48-(8+12)=48-20=28 students

(cosA+ secA)^2+(sinA+ cosecA)^2=(cosA+ 1/ cosA)^2+ ( sinA+ 1/ sinA)^2= ( cos^2A+1/ cosA)^2+( sinA^2+1/sinA)^2=((cosA+1)^2/cos^2A + sin^2A+1/sin^2A)=2-2cos^2A+2sin^2A

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given: BF||AC(i)we know that if Triangles lie on the same base and between the same parallels then their areas are equal.

Triangles ACB and ACF are on the same base AC and between the same parallels AC and BFTherefore, ar(ACB)=ar(ACF) --------(1)hence proved.

(ii)adding ar AEDF to both sides of (1)ar(ACB)+ar(AEDF)=ar(ACF)+ar(AEDF)=>ar(ABCDE)=ar(AEDF)Hence proved.

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(1) 8^-1 * 5^3/2^-4= (2)^[-3+4] *125= 2 *125= 250

(2) (5^-1 * 2^-1)* 6= (1/5 *1/2)*6= 6/10= 3/5

Positive value of cos is taken because only then it satisfies the given equation.

Prove that 2 only

But the ans is 5

9-5b-[2a-b-3a+2c+a-2b]=9-5b-[-3b+2c]=9-5b+3b-2c=9-2b-2c

it is one type fruit and it helps to our health

7 × 11 × 13 + 13

13 ( 7×11 + 1)

Iske do factor hai to ek bhajya sankhya hai.

7 × 6 × 5 × 4 × 3 × 2 × 2 + 5

5 ( 7 × 6 × 4 × 3 × 2 × 2 + 1)

Iske do factor hai to ek bhajya sankhya hai.

(i) (-5)(-5)(-5)/(-5)(-5) = -5

(ii) 125×(5)(5)(5)(5)(5)(5)(5)(6)(6)(6)(6)(6) / ((3×3×3×3×3×10×10×10×10×10) = (125×5×5) = 3125

39. (First digit is second number and second digit is sum)