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2 years ago

I need help please?!!!

Mathematics

2 answers:

poizon [28]2 years ago

8 0

Step-by-step explanation:

$\bigg( \frac{ {7}^{ - 3} }{10} \bigg)^{ - 2} \\ \\ = \bigg( \frac{ 10}{{7}^{ - 3} } \bigg)^{ 2} \\ \\ = \bigg( { 10 \times{7}^{ 3} } \bigg)^{ 2} \\ \\ = \bigg( { 10^{ 2} \times{7}^{ 3 \times 2} } \bigg) \\ \\ = {7}^{ 6} \times 10^{ 2}$

Vadim26 [7]2 years ago

7 0

Answer:

Step-by-step explanation:

$(\frac{7^{-3}}{10})^{-2}=\frac{7^({-3*-2})}{10^{-2}}\\\\=\frac{7^{6}}{10^{-2}}\\\\=7^{6}*10^{2}$

$(a^{m})^{n}=a^{m*n}\\\\\frac{1}{a^{-m}}=a^{m}$

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Putting the value of y'(x) in equation (1)

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$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

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Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

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Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

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Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

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