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KATRIN_1 [288]
3 years ago
6

Expand the expression using the Binomial Theorem and Pascal's Triangle: (3x-1)^3

Mathematics
1 answer:
Readme [11.4K]3 years ago
6 0
(3x)^3+3(3x)^2*-1 +3(3x)(-1)^2+(-1)^3 then simply that down to 27x^3-27x^3+9x-1
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7 0
3 years ago
4y+8x=78<br> solve for the values of x and y
mamaluj [8]
Y=39/2-2X
X=39/4-1/2Y
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3 years ago
2root3 sin^<a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="ac9eec">[email&#160;protected]</a> - <a href
Phantasy [73]

Answer:

Step-by-step explanation:

2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3})  } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}

\alpha =2 n\pi+ \frac{\pi }{6} ,2n\pi +\frac{11\pi }{6} (general~solution)

or~cos\alpha =-\frac{7}{4\sqrt{3} } \\ \alpha =cos^{-1}( \frac{-7}{4\sqrt{3} } )

5 0
3 years ago
Find a basis for the orthogonal complement of the subspace of R4 spanned by the vectors. v1 = (1, 4, -5, 3), v2 = (4, 15, 0, 5),
Arlecino [84]

Answer:

W1 =  ( -75, 20, 1 , 0 )

W2 = ( 25, -7 , 0, 1 )

Step-by-step explanation:

attached  below is the remaining part of the solution

for a homogenous system of equation ; Ax = 0

x1 + 4x2 -5x3 + 3x4 = 0

-x2 + 20x3 -7x4 = 0         note: x3 and x4 are free variables

we can take x3 = 0 and x4 = 1 , hence ; x2 = -7

∴ x1 - 28 + 3 = 0 = x1 = 25

W2 = ( x1 ,x2, x3, x4 ) = ( 25, -7 , 0, 1 )

now lets take x3 = 1  and x4 = 0   hence x2 = 20 , x1 = -75

∴ W1 =  ( x1 , x2 , x3, x4 ) = ( -75, 20, 1 , 0 )

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2 years ago
Determine the rate and the unit rate. $18 for 6 volleyballs
Nadusha1986 [10]

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3.00$ is the answer

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3 years ago
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