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4 years ago

Expand using the identity ( 2x + 3y)^2

Mathematics

2 answers:

drek231 [11]4 years ago

8 0

Hi there!

~

$( 2x + 3y)^2$

$= (2x + 3y) (2x + 3y)\\= (2x)(2x)+ (2x) (3y) + (3y)(2x) +(3y)(3y)\\=4x^2 + 6xy + 6xy + 9y^2\\= 4x^2 + 12xy + 9y^2$

Answer : $4x^2 + 12xy + 9y^2$

Hope this helped you!

kipiarov [429]4 years ago

8 0

Answer:

Identity used:

(a+b)²=a²+b²+2ab

Answer:

 $( {2x + 3y})^{2} \\ = 4 {x}^{2} + {9y}^{2} + 2 \times 2x \times 3y \\ = 4 {x}^{2} + 9 {y}^{2} + 12xy$ 

hope it helped you:)

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Step-by-step explanation:

$\cos\theta =\frac{-2}{5}$

As both $sin\,\theta$ ,

$\theta$ lies in the third quadrant.

In the third quadrant,

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$\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}$

$\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}$

$\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}$

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$\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }$

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