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As they fell short of 2 runs from a double century that means they scored 200-2=198 runsnow sachhin scored doubleLet score by Rahul be xhence Sachin scored 2xhence2x+X=1983x=198x=66Hence Rahul scored 66 runs and Sachin scored 2x=2*66=132 runs

we have to first find 15th odd number after 36first odd is 37 and d=2so 15th is a+14d=37+28=65cube of 65=65×65×65=274625

7.

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1 is the zero of the polynomialso 4(1)^2-8k(1)-9=04-8k-9=0so 8k=-5so k=-5/8

clealry a 2^x -1 is odd so all should possibleA= {1,2,3,5,6,7,8,9}

The lengths of the three pieces of timber are 42 m, 49 m and 63 m.The greatest possible length of each plank will be given by the HCF of 42, 49 and 63.

Firstly, we will find the HCF of 42 and 49 by division method.

∴The HCF of 42 and 49 is 7.Now, we will find the HCF of 7 and 63.​∴The HCF of 7 and 63 is 7.Therefore, HCF of all three numbers is 7Hence, the greatest possible length of each plank is 7 m.

1

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odd number is in form of 2n - 1.so,a(15) = 30 - 1 = 29

_3/_2

_3/_2

29 is the 15th positive odd number.

Suppose the greatest number which can divide 76, 132 and 160 ispand the remainder isr.

Also letq1,q2andq3are the quotients of 76, 132 and 160 respectively.

So we have;

pq1+r=76...(i)pq2+r=132...(ii)pq3+r=160...(iii)

Nowfrom(i),(ii)and(iii)weget;p(q2−q1)=132−76=56p(q3−q2)=160−132=28p(q3−q1)=160−76=84

Therefore HCF of 56, 28 and 84 is given by;

56=2×2×2×728=2×2×784=2×2×3×7

HCFOF56,38and84=2×2×7=28Therefore the required greatest number is 28

the correct answer is 35cm

B as x=2/5 which is not a natural number

5x580/60 = Rs. 48.3keep it up beta

good but not no the answer

correct answer is 40.3

total cost=48.3 Rs. is answer.

Let n be any natural number 1,2,3..........(4)1=4(4)2=16(4)3=64(4)4=256.... therefore 4^n never ends with 0 .

And as it end with 6 and 4 thats why it is divisible by 2

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:If the number 4n, for any n, where to end with digit zero, then it would be divisible by 5. That is the prime factorization of 4n would contain the prime 5. This is not possible because 4n = (2)²n ; so the only prime factorization of 4n is 2. So the uniqueness of the fundamental thm of arithmetic guarantees that there are no other prime in the factorization of 4n. So there is no natural number n for which 4n ends with digit zero.

Let the no be 899*x+63 [dividend=divisor*quotient+remainder]When this no is divided by 29, 899*x gives remainder zero and 63 gives remainder 5 therefore required remainder is 5.

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1. 92/14

= 14*6 = 84

quotient = 6; remainder = 8

First find out the LCM of 20,25,35 and 40

20=2×2×5

25=5×5

35=1×5×7

40=2×`2×2×5

LCM of 20,25,35 & 40 = 2×2×5×5×7×2=1400when divided by 20,25,35,40 leaves a remainder 14,19,29,34i.e 6 less than the divisor in each case

Hence the required number =1400−6=1394

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Given:-i)In ∆PQR,seg.PM is a median.ii)PM=9iii)PQ²+PR²=290.To find:-QR=?Solution:-(By using Appolonius theorem)PQ²+PR²=2PM²+2QM²(Substitute the values)290=2(9)²+2QM²290=2(81)+2QM²290-162=2QM²128=2QM²128/2=QM²64=QM² (By taking sq.root )8=QMi.e.QM=8We know that,PM is median on seg.QR:.QM=1/2(QR)i.e.2QM=QR(Substitute the value)2(8)=QRQR=16

Suppose the greatest number which can divide 76,132 and 160 is p and reminder be r Also the quotient be q1,q2 and q3 for 76,132 and 160 respectively Thus pq1+r=76...........(1) pq2+r=132............... (2) pq3+r=160.............. (3) From 1,2 and 3 we get P(q2-q1)=132-76=56 P(q3-q2)=160-132=28 P(q3-q1)=160-76=84 Therefore the hcf of 56,28 and 84 are 56=2*2*2*7 28=2*2*7 84=2*2*3*7 HCF=2*2*7=28 Therefore the greatest which can divide 76,132 and 160 and leave the same reminder is 28

i can not understand ur HCF method ..Can u plz click picture by solving it in a notebook and snd iy to me

The answer for this question is39.

Explanation :

Step 1-Take LCM (9,10,15) = 90

Step 2 - Divide 1936 by 90.

1936÷90 = 21 (quotient) and 46(remainder)

Step 3 - To get 7 as remainder you must subtract 39.

As 46–39 = 7

Result :1936–39 = 1897

1897÷9 = 210(quotient) + 7(remainder)

1897÷10 = 189(quotient) + 7(remainder)

1897÷15 = 126(quotient) + 7(remainder

Disjoint of sets using Venn diagram is shown by two non-overlapping closed regions and said inclusions are shown by showing one closed curve lying entirely within another.

Two sets A and B are said to be disjoint, if they have no element in common.Thus, A = {1, 2, 3} and B = {5, 7, 9} are disjoint sets; but the sets C = {3, 5, 7} and D = {7, 9, 11} are not disjoint; for, 7 is the common element of A and B.

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