answer this work sheet
Answers
Answer:
so many questions !!!
I answer a few, not all
Q9. Pythagorean triplet
we all know that P2 + B2 = H2 are called pythagorean triplets. however. we need to know two numbers to find the third one in this format. so a relation has been established between the triplets using one variable, so that we may be able to find the triplet if only one number is known.
the triplets are
m^2 - 1, 2m and m^2 + 1
however, it should be kept in mind that this method will give us atleast one of the many possible triplets. this method cannot be used to find all the triplets containing a particular number
given one of the triplets is 12
now we see, which one it is. since we don't know, we will equate it to each of the triplets to see to which it best suits
so m^2-1 = 12
m^2 = 13
m = √13 (not an integer)
let 2m = 12
m = 6
this gives one set of triplets
6^2-1, 2*6, 6^2+1
= 35, 6, 37
(note: it did not give 5, 12, 13)
Q10 . we may use division method of finding square root of 8934. the remainder that we will get in the process will be our answer as the same number will need to be subtracted from the given number to make it a perfect square.
9 |. 8934. | 94
+9. |.-81. |
----------------
184 |. 834
4 |. - 736
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98
Hence, our answer is 98. By subtracting this number from 8934. the number will become perfect square whose square root will be 94.
Q14
(2x+5y)^2 + (2x-5y)^2
we will use the identity
a^2 + b^2 = (a+b)^2 - 2ab
a^2 - b^2 = (a+b)(a-b)
the problem becomes
(2x+5y+2x-5y)^2 - 2(2x+5y)(2x-5y)
(4x)^2- 2((2x)^2 - (5y)^2)
16x^2 - 2(4x^2 - 25y^2)
16x^2 - 8x^2 + 50y^2
8x^2 + 50y^2
Q15
9^3 x 27 x t^4
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3^2 x 3^4 x t^2
(3^2)^3 x 3^3 x t^4
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3^2 x 3^4 x t^2
3^6 x 3^3 x t^4
----------------------
3^2 x 3^4 x t^2
3^9 x t^4
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3^6 x t^2
3^(9-6) x t^(4-2)
3^3 x t^2
= 27t^2