Question

6.87 x 6.87 +2x 6.87 x 3.13+ 3.13 x 3.13 = ?
A) 100
C) 10.2
B)9.82
C)10.2
D)None of these

how to solve?
please help me​

Answer 1

Answer:

Correct Answer: C

Solution :

The number of selections = coefficient of \[{{x}^{8}}\]in \[(1+x+{{x}^{2}}+.........+{{x}^{8}})(1+x+{{x}^{2}}+......+{{x}^{8}}).{{(1+x)}^{8}}\]= coefficient of \[{{x}^{8}}\] in \[{{\frac{(1-{{x}^{9}})}{{{(1-x)}^{2}}}}^{2}}{{(1+x)}^{8}}\] = coefficient of \[{{x}^{8}}\] in \[{{(1+x)}^{8}}{{(1-x)}^{-2}}\] = coefficient of \[{{x}^{8}}\] in \[{{(}^{8}}{{C}_{0}}{{+}^{8}}{{C}_{1}}x{{+}^{8}}{{C}_{2}}{{x}^{2}}+.......+{{\,}^{8}}{{C}_{8}}{{x}^{8}})\] \[\times (1+2x+3{{x}^{2}}+4{{x}^{3}}+......+9{{x}^{8}}+.....)\] \[=9\ .{{\ }^{8}}{{C}_{0}}+8\ .{{\ }^{8}}{{C}_{1}}+7\ .{{\ }^{8}}{{C}_{2}}+.........+1\ .{{\ }^{8}}{{C}_{8}}\] \[={{C}_{0}}+2{{C}_{1}}+3{{C}_{2}}+.....+9{{C}_{8}}\] \[[{{C}_{r}}{{=}^{8}}{{C}_{r}}]\] Now \[{{C}_{0}}x+{{C}_{1}}{{x}^{2}}+.......+{{C}_{8}}{{x}^{9}}=x{{(1+x)}^{8}}\] Differentiating with respect to \[x\], we get \[{{C}_{0}}+2{{C}_{1}}x+3{{C}_{2}}{{x}^{2}}+...9{{C}_{8}}{{x}^{8}}={{(1+x)}^{8}}+8x{{(1+x)}^{7}}\] Putting \[x=1,\ \]we get \[{{C}_{0}}+2{{C}_{1}}+3{{C}_{2}}+......+9{{C}_{8}}\] \[={{2}^{8}}+8\ .\ {{2}^{7}}={{2}^{7}}.(2+8)=10\ .\ {{2}^{7}}\].

Step-by-step explanation:

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Answer 2

Answer:

(A) 100

Step-by-step explanation:

6.87*6.87 =47.1969

2*6.87 *3.13= 43.0062

3.13*3.13 = 9.7969