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Using Euclid's division lemma -

100 = (60× 1) + 40

60 = (40 × 1) + 20

40 = (20 × 2) + 0

Now, the remainder is 0. So, the HCF of 60 and 100 is 20

Use Euclid's division lemma to show that any positive odd integer is of the form 6q+1,6q+3,6q+5 where q is a certain integer.

Answer:-Let a be a positive odd integer a=bq+rb=6 a=6q+r, 0≤r<6. So,the possible values of r are 0,1,2,3,4,5

Set of positive odd integers are {1,3,5,7,9......}put a=1,3,5,7,9......a=bq+r1=6(0)+1=6q+1 [r=1]3=6(0)+3=6q+3 [r=3]5=6(0)+5=6q+5 [r=5]7=6(1)+1=6q+1 [r=1]9=6(1)+3=6q+3 [r=3]

So,any positive integer is of the form 6q+1,6q+3,6q+5 where q is certain integer.

tq

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solution

diagram

csa of cube=4×4×4=64cm2tsa of cube=6×(4^2)=96cm2csa of cuboid=2(lh+bh)2×(40+30)140cm2tsa of cuboid=2(lh+bh+lb)2(118)236cm2

236 cm*2 is the best answer

236 CM 2 is best answer

1) LSA of a cube: 4a^2 4*4*4 64cmTSA of a cube: 6a^2 6*4*4 96cm2) TSA of a cubiod: 2(lb+bh+hl) 2(8*6+6*5+5*8) 2(48+30+40) 2*118 236cmLSA of a cubiod: 2h(l+b) 2*5(8+6) 10*14 140cm

Power P= 120 Watt

Time t= 20 minutes= 20× 60 s= 1200 seconds

We know that,

Power P = Work W ( or Energy E)/ Time taken t

Then, Energy E= Power P × Time t

So, Energy E= 120 W× 1200= 144000 J or 144 KJ.

6 is the cuberoot of larger number.

6 is the cuberoot of larger number

6 is the cube root of largest number

Let, the first find the cube root of 3. I:e 27Now add 27+189=216hence the cube root of 216 is 6

6 will be the cube root of the largest no.

smallest natural number is 44

then it will become 1188/44 =27

and 27 is perfect cube of 3

smallest number is 44

then it will become 1188/44 = 27

and 27 is perfect cube of 3 regard

perfect cubes are 1,8,27,64 so probability=4/100=1/252 digit divisible by 4 12,16,..,96 so 22probability=22/100=11/50

68600=2*2*2*5*5*7*7*7

So68600isnotaperfectcube

Sowehavetomultiple5togetgetperfectcubethenwewillget343000

68600 is the correct answer

68600 is the correct ans

68600 is the correct answer of the given question

68600 is the correct answer

68600 is the right answer

The only non-perfect cube in question number 20 is 243.

a

On factorising 243 into prime factors, we get:243=3×3×3×3×3

On grouping the factors in triples of equal factors, we get:243={3×3×3}×3×3

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is multiplied by 3, the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be multiplied by 3 to make it a perfect cube.

b

On factorising 243 into prime factors, we get:243=3×3×3×3×3

On grouping the factors in triples of equal factors, we get:243={3×3×3}×3×3

It is evident that the prime factors of 243 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 243 is not a perfect cube. However, if the number is divided by (3×3=9), the factors can be grouped into triples of equal factors such that no factor is left over.

Thus, 243 should be divided by 9 to make it a perfect cube.

16384÷4=40964096 is the cube of 16

prime factorization of 1863400 is (2^3×5^2×7×11^3) so to make this no a perfect cube we have to divide it by 5×5 and 7 as now 10648 will come which is cube of 22

bhai ji thanks.Scholar ho yaar ap

Vegetative reproduction is any form of asexual reproduction occurring in plants in which a new plant grows from a fragment of the parent plant or grows from a specialized reproductive structure. Plant Propagation is the process of plant reproduction of a species or cultivar, and it can be sexual or asexual.

The degree of a polynomial is the highest degree of its monomials with non-zero coefficients.

Therefore, for polynomial-3z^5 + 2z^4 + 4z^3 + z +1Degree is 5

In Euclidean geometry, a parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

Area:base × height

secA/tanA = (1/cosA)/(sinA/cosA) = 1/sinA

In mathematics, an "identity" is an equation which is always true. These can be "trivially" true, like "x=x" or usefully true, such as the Pythagorean Theorem's "a2+b2=c2" for right triangles. There are loads of trigonometric identities, but the following are the ones you're most likely to see and use.

Basic and Pythagorean Identities

\csc(x) = \dfrac{1}{\sin(x)}csc(x)=sin(x)1​

\sin(x) = \dfrac{1}{\csc(x)}sin(x)=csc(x)1​

\sec(x) = \dfrac{1}{\cos(x)}sec(x)=cos(x)1​

\cos(x) = \dfrac{1}{\sec(x)}cos(x)=sec(x)1​

\cot(x) = \dfrac{1}{\tan(x)} = \dfrac{\cos(x)}{\sin(x)}cot(x)=tan(x)1​=sin(x)cos(x)​

\tan(x) = \dfrac{1}{\cot(x)} = \dfrac{\sin(x)}{\cos(x)}tan(x)=cot(x)1​=cos(x)sin(x)​

Notice how a "co-(something)" trig ratio is always the reciprocal of some "non-co" ratio. You can use this fact to help you keep straight that cosecant goes with sine and secant goes with cosine.

The following (particularly the first of the three below) are called "Pythagorean" identities.

sin2(t) + cos2(t) = 1

tan2(t) + 1 = sec2(t)

1 + cot2(t) = csc2(t)

Note that the three identities above all involve squaring and the number1. You can see the Pythagorean-Thereom relationship clearly if you considerthe unit circle, where the angle ist, the "opposite" side issin(t) =y,the "adjacent" side iscos(t) =x, and the hypotenuse is1.

4 termsax^3+bx^3+cx+d

b...Themean value theoremstates that p/(x) will have 3 roots. However p/(x) will have degree 2 and therefore will have at most 2 roots. So we get a contra- diction, which means p(x) cannot exist. Thus a polynomial of degree 3 has at most three real roots

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what is the 2nd step? I can't understand

I have replaced them with π/4

polynomial 2x-1to prove let polynomial be ax+bnowwhenx=1ax+b=1 that is a+b=1....1and when X=2ax+b=3 that is ,2a+b=3....2hence subtracting both the equation-a=-2 hence a =2put in equation 1hence b=-1hence polynomial will be 2x-1

The no. of terms in a polynomial of degree 6. is 6+1 = 7

cot (85°) + cos(75°)

Cot 85°+Cos75°

Use formula

cot(90°- A)=tan(A) cos(90°- A)= sin(A)

cot(90°-5°)+cos(90°-15°)

tan 5° + sin 15°

Euclid axiomsonly one line can be drawn through a point parallel to another linea straight line can be drawn between any two points

angle DAC = angle I + 3 and angle Bac = 2+4 . here I = 2 and 3 = 4 So give a Common name for 1 and 2 is x and for 3 and 4 is y. So I + 3 = X + Y. and 2 + 4 = X+\So both is Same

Fifth postulate of Euclidgeometry: If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles

energy consumed= 1500*101500015kw-h

because air contains moisture this moisture can cause the setting of plaster of Paris into hard mass gypsum.