percentage of increase =36% value of increase =611.2N/mm original tensile strength
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Answered by
2
The first equation given,
x-\frac{1}{x}=5x−
x
1
=5 …….. (i)
If we do the cube of equation (i) we get,
\left(x-\frac{1}{x}\right)^{3}=5^{3}(x−
x
1
)
3
=5
3
x^{3}-\left(\frac{1}{x}\right)^{3}-3 \times x \times\left(\frac{1}{x}\right) \times\left(x-\frac{1}{x}\right)=125x
3
−(
x
1
)
3
−3×x×(
x
1
)×(x−
x
1
)=125
x^{3}-\left(\frac{1}{x}\right)^{3}-3\times(x-\frac{1}{x})=125x
3
−(
x
1
)
3
−3×(x−
x
1
)=125
x^{3}-\left(\frac{1}{x}\right)^{3}-3 \times 5=125x
3
−(
x
1
)
3
−3×5=125
Grouping the terms,
x^{3}-\left(\frac{1}{x}\right)^{3}=125+15x
3
−(
x
1
)
3
=125+15
x^{3}-\left(\frac{1}{x}\right)^{3}=140x
3
−(
x
1
)
3
=140
Answered by
0
Answer:
Step-by-step explanation:
yes
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