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In triangle PQR and XYZ

PQ=XY PR=XZ QR=YZ

By SSS rule the triangles are congruent

in

4/15 is the best answer for the question

2x²+7x²-3x-18= 9x²-3x-18= 3(3x²-x-6)

2x+7x=-39x=-3x=-3/9x=-3

the number of jugs 18÷(3/5)=18*5/3=30

thanks

We know when a + b + c = 0then a³ + b³ + c³ = 3abc

Use this identity

So, here we have (-12)³ + 4³ + 8³

Observe (-12) + 4 + 8 = 0

So, (-12)³ + 4³ + 8³

= 3 × (-12) × 4 × 8

= -1125

3.20 litres of water is left into bucket

I 🤔 think, as

5.75 litres of water was already present in the bucket

10.25 litres of water was added later to the 5.75 litresso, 10.25 + 5.75 = 16.00 litres at last Srikant drew out 12.80 litres from 16 litresso, 16.00 - 12.80 = 3.20 litres

3.20 litres of water is left into buckets

3.20litres of water is left into buckets

3.20 litters is the right answer of the following

3.20 l is the correct answer of the given question is

Let the capacity of bucket x..80x/100 - 200x/300 = 240x=600X= 15capacity is of 15 litre

arranging 15 numbers in ascending order 1,1,1,1,2,2,2,3,3,4,5,5,5,6,6 so median is 8th term = 3 here 1 is appearing 4 times which is maximum so mode = 1

tan^6A + 3tan^2A.sec^2A + 1 = tan^6A + 3tan^2A(1+ tan^2A ) + 1 [since sec^2A = 1+ tan^2A] = tan^6A + 3tan^4A + 3 tan^2A + 1 = (tan^2A)^3+ 3(tan^2A)^2+ 3 (tan^2A) + 1It is in the form of a^3+ 3a^2b + 3ab^2+b^3= (a + b)^3∴ (tan^2A)^3+ 3(tan^2A)^2+ 3 (tan^2A) + 1 = (tan^2A + 1)^3 = (sec^2A)^3 = sec^6A

zero of p(x) is 5 is the p(x)of thos question

Since 1 and 3 are the zeros or roots of the given polynomial2x4- 7x3- 13x2+ 63x - 45, use the synthetic division method and factorize2x4- 7x3- 13x2+ 63x - 45 by dividing it with (x- 1) to get the remainder polynomial as 2x3- 5x2- 18x + 45.Now again divide the polynomial2x3- 5x2- 18x + 45 by (x - 3) to get the remainder as 2x2+ x - 15.Since the constant term in the equation2x2+ x - 15 is 15, it will be divisible by±3 ,±5, or± 15Use synthetic division again on the equation2x2+ x - 15 and factorize it by -3 to get the remainder as 2x - 5.Now, 2x - 5 has the root as 5/2.So, the zeros of2x4- 7x3- 13x2+ 63x - 45 are 1, 3,-3,5/2

1)Force can be defined as a push or a pull that tends to change the state of an object or changes the direction

2)Forcesare due to an interaction of at least two objects. It may change the state of motion of an object. ... If the twoforcesacting on object are equal in magnitude but opposite in direction, then the netforceacting on the body is zero.

78×82(80-2)(80+2)(a+b)(a-b)=a^2-b^2(80-2)(80+2)=(80^2)-(2^2) =6400-4=6396

104×96(100+4)(100-4)100²-4²10000-169984

9984 the correct answer of the given question

9984 is the correct answer

9984 is correct answer

97^3= (100 - 3)^3

Using identity (a − b)^3 = a^3 - b^3 - 3ab(a - b)

(100 - 3)^3= (100)^3 - (3)^3 - 3*100*3(100-3)= 1000000 - 27 - 900*97= 1000000 - 27 - 87300= 1000000 - 87327= 912673

we have changed 54 second Into minutes by multiplying by 5/18

take log in both the sideslogy=xlogxdifferenciate both the sides1/ydy/dx=x/x+logxso dy/dx=y*(1+logx)=x^x*(1+logx)

wrong

Rationalize the denominator

= [(3 + 2i sinθ)(1 + 2i sinθ)] / [(1 - 2i sinθ)(1 + 2i sinθ)]

= (3 + 6i sinθ + 2i sinθ - 4sin^2θ) / (1 + 4sin2θ)

= (3 - 4sin2θ /1 + 4sin^2θ) (8i sinθ /1 + 4sin^2θ)

Given: the complex numbers are real. Therefore

8i sinθ/1 + 4sin^2θ=0

i.e; sinθ = 0

Thus θ = nπ, n∈ Z

The duration of the play i.e. the time for which the play is played is 2 hours and 35 minutes.

(sinθ - cosθ +1 )/(sinθ +cosθ -1)

dividing numerator and denominator by cosθ

[(sinθ - cosθ +1 )cosθ]/[(sinθ +cosθ -1)/cosθ]=(tanθ -1 + secθ )/(tanθ +1 - secθ)=(tanθ + secθ -1)/(tanθ - secθ+1)

As, sec²θ- tan²θ = 1(secθ -tanθ)(secθ +tanθ) = 1

putting this in numerator,[(tanθ + secθ -(sec²θ- tan²θ)]/(tanθ - secθ+1)=[(tanθ + secθ) -(secθ- tanθ)(secθ+tanθ)]/(tanθ - secθ+1)=(tanθ+secθ)[1- (secθ - tanθ)]/(tanθ - secθ+1)=(tanθ+secθ)[1- secθ + tanθ)]/(tanθ - secθ+1)=(tanθ+secθ)

Now, multiplying and dividing by(secθ- tanθ)[(tanθ+secθ)×(secθ- tanθ)]/(secθ- tanθ)=(sec²θ- tan²θ)/(secθ- tanθ)= 1/(secθ- tanθ)=RHS

Hence proved

Number of mugs = 18/1.5 = 12

green is the correct answer of the given question

30-25+22=25+2027=4527/45Baki aage app khud Bana lo ge

correct answer is green

correct answer is green

green is the answer of the following

Given:- the bucket contains 18 litres of the substance.

∴No. of mugs of 1(1/2) litres that can be taken out=18 litres/ 3/2 litres=12 mugs

like me and I will like you

total volume of water in the bucket=

21.25

volume of a mug = 1.25

acq

no. of such mug can be filled = 21.25/1.25

= 17

total volume of water in bucket =21.25volume of one mug =1.25now, 21.25/1.25=17

answer of this question is 17 mugs

17 mugs is correct answer

Like my answer if you find it useful!

Capacity of mug (Actual quantity of milk) =πr2h - 2/3πr3=πr2(h - 2r/3)r = 7/2 and h = 14Capacity of mug = 2695/6 cm3Amount dairy owner B should charge for one mug of milk= 2695/6× 80/1000= Rs. 35.93Value exhibited by dairy owner B: honesty

dude this is the format of the sum..

Perimeter of a triangle= Sum of all sides=42+38+25=105 m

pie*r*r=16pieso r=4so circumference 2pie*r=8pie

16x²-81=(4x²)²-(3²)²Since a²-b²=(a+b)(a-b)

(4x²)²-(3²)²=(4x²+9)(4x²-9)=(4x²+9)(2x+3)(2x-3)

can show thata2+ b2= c2usingAlgebra:-

Take a look at this diagram ... it has that "abc" triangle in it (four of them actually):

Area of Whole Square

It is a big square, with each side having a length ofa+b, so thetotal areais:

A = (a+b)(a+b)

Area of The Pieces

Now let's add up the areas of all the smaller pieces:

First, the smaller (tilted) square has an area of:c2

Each of the four triangles has an area of:ab2

So all four of them together is:4ab2= 2ab

Adding up the tilted square and the 4 triangles gives:A = c2+ 2ab

Both Areas Must Be Equal

The area of thelarge squareis equal to the area of thetilted square and the 4 triangles. This can be written as:

(a+b)(a+b) = c2+ 2ab

NOW, let us rearrange this to see if we can get the pythagoras theorem:

Start with:(a+b)(a+b) = c2+ 2ab

Expand (a+b)(a+b):a2+ 2ab + b2= c2+ 2ab

Subtract "2ab" from both sides:a2+ b2= c2