Maths Help (1 Viewer)

Farhanthestudent005 · Sep 13, 2021

Will appreciate if someone helps me with these. Thanks

Drongoski · Sep 13, 2021

Lith_30 · Sep 13, 2021

for question 5 we have to find the general term of $(3x-y)^{10}$

$\text{General Term}\\\\{10 \choose k}{\cdot}(-1)^{10-k}{\cdot}y^{10-k}{\cdot}3^k{\cdot}x^k$

for x to be of power 7, k has to equal 7 so we sub it into the general term

$\\{10 \choose 7}{\cdot}(-1)^{10-7}{\cdot}y^{10-7}{\cdot}3^7{\cdot}x^7\\\\=-262440y^3x^7$

For question 6 we have to expand the brackets and then simplify

$\\(2-\sqrt{5})^4\\\\=2^4+4(2^3)(-\sqrt{5})+6(2^2)(-\sqrt{5})^2+4(2)(-\sqrt{5})^3+(-\sqrt{5})^4\\\\=16-32\sqrt{5}+120-40\sqrt{5}+25\\\\=161-72\sqrt{5}$

Therefore $a=161$ and $b=-72$

Farhanthestudent005 · Sep 14, 2021

Lith_30 said:
for question 5 we have to find the general term of $(3x-y)^{10}$

$\text{General Term}\\\\{10 \choose k}{\cdot}(-1)^{10-k}{\cdot}y^{10-k}{\cdot}3^k{\cdot}x^k$

for x to be of power 7, k has to equal 7 so we sub it into the general term

$\\{10 \choose 7}{\cdot}(-1)^{10-7}{\cdot}y^{10-7}{\cdot}3^7{\cdot}x^7\\\\=-262440y^3x^7$

For question 6 we have to expand the brackets and then simplify

$\\(2-\sqrt{5})^4\\\\=2^4+4(2^3)(-\sqrt{5})+6(2^2)(-\sqrt{5})^2+4(2)(-\sqrt{5})^3+(-\sqrt{5})^4\\\\=16-32\sqrt{5}+120-40\sqrt{5}+25\\\\=161-72\sqrt{5}$

Therefore $a=161$ and $b=-72$

Thank you.. Appreciate it

Maths Help (1 Viewer)

Farhanthestudent005

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Drongoski

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Lith_30

o_o

Farhanthestudent005

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