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Magnetic charge” (i.e. the magnetic counterpart of electric charge) is actually calledpole strength. In SI units, the units depend on how pole strength is defined. If defined as the force per unitBfield, the SI units are ampere-meter^2(A m^2). If defined as the force per unitHfield, the SI units are webers (Wb).

In CGS Gaussian units, theBandHfields have the same units, and pole strength has units ofunit poles.

Circumference of circle = 2 × π × r =>2 × 22/7 × 154/10=>(2×22×154)/7×10=>96.8cm

Hence

Circumference of circle = 96.8cm

Now

perimeter of two semicircle = 2 × perimeter of semi circle

=>2×perimeter of semi circle =>2× (π ×r)=>2×22/7×154/10=>96.8cm

243.80+ 60.00+ 8.28=312.08- 14.00=298.08

a zero of a polynomial is a value of x which makes P(x)=0

so, for P(x)=x+5, you find the zero00s by setting equal to zero and solving for x:

0=x+5

x=-5

P(-5) = (-5)+5 = 0.

this is a polynomial of the power 1which has one zero that is x=-5if we put it in the equation our polynomial will be 0

okay

Given: AM is the median of ∆ABC & PN is the median of ∆PQR.

AB = PQ, BC = QR & AM = PN

To Show: (i) ΔABM ≅ ΔPQN(ii) ΔABC ≅ ΔPQR

Proof: Since AM & PN is the median of ∆ABC

(i)1/2 BC = BM & 1/2QR = QN (AM and PN are median) Now, BC = QR. (given)

⇒ 1/2 BC = 1/2QR (Divide both sides by 2) ⇒ BM = QN

In ΔABM and ΔPQN, AM = PN (Given) AB = PQ (Given) BM = QN (Proved above)

Therefore, ΔABM ≅ ΔPQN (by SSS congruence rule) ∠B = ∠Q (CPCT)

(ii)In ΔABC & ΔPQR,

AB = PQ (Given) ∠B = ∠Q(proved above in part i)BC = QR (Given) Therefore, ΔABC ≅ ΔPQR ( by SAS congruence rule)

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put x = 4, 3×4 -5 -7 12 -12 = 0yes it is zero

thanks friend

First Axiom: Things which are equal to the same thing are also equal to one another.

Second Axiom: If equals are added to equals, the whole are equal.

Third Axiom: If equals be subtracted from equals, the remainders are equal.

Fourth Axiom: Things which coincide with one another are equal to one another.

Fifth Axiom: The whole is greater than the part.

X - 15 = 25On adding 15 to both sides ,we havex- 15 + 15 = 25 + 15ie, x = 40 (using Euclid's second axiom)

thanks

p (x)= x^2-8x+kp (x)=Ax^2+Bx+C(The equation is inthis form)

Let the zeroes be 'a' and 'b'

It is given that=> a^2+b^2=40=> (a+b)^2 - 2ab = 40 ----1

sum of zeroes=> a+b = -B/A = -(-8)/1 = 8

Product of zeroes => a*b = C/A = k

Substitute these values in the equation1 we get=> 8^2 - 2k = 40=> 64 - 2k = 40=> 2k = 24=> k = 12

thank uu so much

As given,Quotient = x^2 + 2x - 3,Remainder = 5

Then,p(x) = (x^2 + 2x - 3)*(3x - 1) + 5= 3x^3 - x^2 + 6x^2 - 2x - 9x + 3 + 5= 3x^3 + 5x^2 - 11x + 8

Therefore, p(x) = 3x^3 + 5x^2 - 11x + 8

If 3, - 1, and - 1/3

Are zeros of polynomial 3x^3 - 5x^2 - 11x - 3

Polynomial is of form ax^3 + bx^2 + cx + d

Then, sum of zeros= 3 + (-1) + (-1/3) = 5/3= - b/a

Sum of product of pair of zeroes= 3*-1 + - 1*-1/3 + 3*-1/3= - 3 + 1/3 - 1= - 4 + 1/3 = - 11/3 = c/a

Product of zeroes= 3*-1*-1/3 = 1= - d/a

Similarly, solve others

SP = 750CP = 600

Selling price is greater than cost price.ao there will be profit.

profit percentage = {(SP-CP)/CP} ×100

i.e., {(750-600)/600} ×100= (150/600)×100 =100/4 = 25%

therefore profit percentage is 25%

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Q. 2

State examples of Euclid's axioms in our daily life.

Answer

Take the 5 Euclid’s axioms one by one –

Axiom 1: Things which are equal to the same thing are also equal to one another.

Example: Take a simple example.

Say, Raj, Megh, and Anand are school friends. Raj gets marks equal to Megh’s and Anand gets marks equal to Megh’s; so by the first axiom, Raj and Anand’s marks are also equal to one another.

Axiom 2: If equals are added to equals, the whole is equal.

Example: Say, Karan and Simran are artists and they buy the same set of paint consisting of 5 colors. This means, Karan has 5 color set and Simran to has same 5 color set.

⇒(Karan’s 5 colour set) = (Simran’s 5 colour set) …(i)

Now, they need the special set of 3 paint brushes each and so they buy it. This means Karan buys 3 paint brushes and Simran too buy the same kind of 3 paint brushes.

⇒(Karan’s 3 paint brushes) = (Simran’s 3 paint brushes) …(ii)

Add the two equals (i) and (ii), we get

(Karan’s 5 color set) + (Karan’s 3 paint brushes) = (Simran’s 5 colour set) + (Simran’s 3 paint brushes)

⇒Karan’s 8 paint items = Simran’s 8 paint items

This well proves that even after adding their equals, the whole were equal.

A

1. sun rises in East2.two parallel lines never meet each other

8Daily Life Examples Of Axioms

0 is a Natural Number. 0 is a natural number, which is accepted by allthepeople on earth. ...

Sun Rises InTheEast. ...

God is one. ...

TwoParallel Lines Never Intersect Each Other. ...

India is a Part of Asia. ...

Probability lies between 0 to 1. ...

TheEarth turns 360 DegreesEveryday. ...

All planets Revolve aroundtheSun.

1. sun rises in East2. two parallel lines never meet each other

1/√7=1/2.645= 0.378

Sum of zeroes is -b/a=(5/2)hence option c

4x*x*x + x*x + x + 6 - ( 2x*x*x - 4x*x + 3x + 5 ) = 4x*x*x + x*x + x + 6 - 2x*x*x + 4x*x - 3x - 5 = 2x*x*x + 5x*x - 2x + 1

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let x = (0.0076)1/5log x= log (0.0076)1/5=1/5 log ( 0.0076)

bar= 1/5 ( 3.8808)bar= 1.5762then antilogx= 0.3769...

answer is answeri..not asking me again

α² +β² can be written as (α +β)² - 2αβ

p(x) = 2x² - 5x + 7a = 2 , b =- 5 , c = 7

α andβ are the zeros of p(x)

we know that ,sum of zeros =α +β = -b/a = 5/2

product of zeros = c/a = 7/2

2α + 3β and 3α + 2β are zeros of a polynomial.

sum of zeros = 2α + 3β+ 3α + 2β = 5α + 5β = 5 [α +β] = 5× 5/2 = 25/2

product of zeros = (2α + 3β)(3α + 2β) = 2α [3α + 2β] + 3β [3α + 2β] = 6α² + 4αβ + 9αβ + 6β² = 6α² + 13αβ +6β² = 6 [α² +β² ] + 13αβ = 6 [ (α +β)² - 2αβ ] + 13αβ = 6 [ ( 5/2)² - 2× 7/2 ] + 13× 7/2 = 6 [ 25/4 - 7 ] + 91/2 = 6 [ 25/4 - 28/4 ] + 91/2 = 6 [ -3/4 ] + 91/2 = -18/4 + 91/2 = -9/2 + 91/2 = 82/2 = 41

-18/4 = -9/2 [ simplest form ]

a quadratic polynomial is given by:-

k { x² - (sum of zeros)x + (product of zeros) }

k {x² - 5/2x + 41}

k = 2

2 {x² - 5/2x + 41 ]

2x² - 5x + 82 -----> is the required polynomial

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let x = (0.0076)1/5log x= log (0.0076)1/5=1/5 log ( 0.0076)

bar= 1/5 ( 3.8808)bar= 1.5762then antilogx= 0.3769...

40 runs in 5 innings nowso 40/5 =8 runs

the correct answer is 8 runs

Average in 4 innings = 40Total runs in 4 innings =160 (40*4)Run in 5th inning= 0Total run after 5th inning= 160+0 = 160Average after 5th inning = 160/5 =32

The ‘square root’ of 54756 by long division method is shown in below attached image.

Hence, by dividing 54756 by 2 by long division method, we get the final answer as 234.

The ‘long division method’ is used dividing a “large number (three digits or more)” by a “two-digit (or more) number”. It is set out in a ‘similar way’ to ‘short division’.

Using Remainder theorem:

460 = Divisor x Quotient + Remainder

Let divisor be D, quotient be Q, and Remainder is given 1

So,

460 = Q.D + 1

Q.D = 459.....(1)

Now we have,

Q = 5D + 6....(2)

Solving (1) and (2) we get

Q = 459/D

Putting this in (2) we get

459/D = 5D + 6

459 = 5D² + 6D

5D² + 6D -459 = 0

5D² - 45D + 51D -459 = 0

5D(D - 9) + 51(D - 9) = 0

(5D + 51)(D - 9) = 0

D = 9 or -51/5

As it is given that divisor is natural number, D will only be 9

Divisor = 9

Quotient = 459/9 = 51

thanks

given eⁿ and e-ⁿ are the roots

so, according to the equation product of roots = 2a/3 => eⁿ * e-ⁿ = 2a/3 => 2a/3 = 1 => a = 3/2

sum of roots should be (a+b)/3 => (a+b)/3 = eⁿ +e-ⁿ=> (a+b) = 3( eⁿ +e-ⁿ)

also, the minimum +ve value of a function x+1/x = 2

so , (a+b) = 3*2 => b = 6-a = 9/2 ... the least intergral value is > 9/2. = 5

option B

Given, radius of the cone, R = 2r

Slant height of cone, L =l/2

Total surface area of cone =πR (L + R)

=π ×2r [ (l/ 2) + 2r ]

=π ×2r [ (l+ 4r) / 2 ]

=πr [l+ 4r ]

Let two parts be x and y

so x+y=27

A2q,

1/x+1/y=3/20

After putting x+y=27

y=180/x

putting y=180/x in eq x+y=27

x^2-27x+180

after solving

x=15 & 12.

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x^3-3x^2+5x-1-(2x^3+x^2-4x+2)x^3-3x^2+5x-1-2x^3-x^2+4x-2-x^3-4x^2+9x-3

2x = 78x = 39R = (39/3)+7R= 13 + 7R= 20 years

1 1/33*1+1/3=3+1/3=4/3

i) 1.51618ii) 0.9918iii) 0.61836

If 4 is the root we will put x= 4 in the equation 16-36+k= 0 now k = 20 so equation will be x^2-9x+20x^2-(5+4)x+20= 0!x(x-5)-4(x-5)= 0x= 5 and x= 4 are two rootsplease like the solution 👍 ✔️👍