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1 year ago

6

Frank wants to build a garden with the dimensions below.

2 answers:

dexar [7]1 year ago

7 0

Given:
Length = x + x + 3 = 2x + 3
Width = 2 + x

Area = length * width
91 ft² = (2x + 3) (2+x)
91 = 4x + 2x² + 6 + 3x
0 = 2x² + 7x + 6 - 91
0 = 2x² + 7x - 85

(2x + 17) ( x - 5)

x = -17/2 or x = 5

Let x = 5 ;

length = 2x + 3 = 2(5) + 3 = 13
width = 2 + x = 2 + 5 = 7

Area = 13 x 7 = 91

Perimeter = 2(length + width)
Perimeter = 2(13 + 7)
Perimeter = 2(20)
Perimeter = 20 feet of fencing

lozanna [386]1 year ago

7 0

Answer:

The length is 13 and the width is 7, but the length of fencing, is 40, because there are 4 sides. Two thirteens = 26 + Two sevens = 14

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lord [1]

I suppose

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Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

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Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

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The coefficient matrix is non-singular, so it has an inverse. Multiplying both sides by that inverse gives

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}-\dfrac{6a-11b+19c}3\\\dfrac{3a-5b+2c}3\\\dfrac{15a-26b+46c}3\end{bmatrix}$

so the vectors do span $P_2$ .

The vectors comprising $H$ form a basis for it because they are linearly independent.

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