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Dahasolnce [82]
3 years ago
8

1.prove that the following three functions are linearly dependent

Mathematics
1 answer:
Setler79 [48]3 years ago
4 0

Proof:

Given any functions f_{1}(x),f_{2}(x),f_{3}(x) they are linearly dependent if we can find values of c_{1},c_{2},c_{3} such that

c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0

Using the given functions in the above equation we get

c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0\\\\c_{1}x^{2}+c_{2}(1-x^{2})+c_{3}(2+x^{2})=0\\\\\Rightarrow (c_{1}-c_{2}+c_{3})x^{2}+c_{1}+c_{2}+2c_{3}=0

This will be satisfied if and only if

c_1-c_2+c_3=0,c_1+c_2+2c_3=0

Solving the equations we get

c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0

Since we have 3 variables and 2 equations thus we will get many solutions  

one being if we put c_3=1 we get

c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0\\\\c_1=\frac{-3}{2}\\\\c_2=\frac{-1}{2}

Thus we have c_1=\frac{-3}{2},c_2=\frac{-1}{2},c_3=1 as one solution. Hence the given functions are linearly dependent.

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Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy
Margaret [11]

Answer:

Part 1) The exact solutions are

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})   and  (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Part 2) (1.79, 8.58)

Step-by-step explanation:

we have

y=x^{2} +3x ----> equation A

y=2x+5 ----> equation B

we know that

When solving the system of equations by graphing, the solution of the system is the intersection points both graphs

<em>Find the exact solutions of the system</em>

equate equation A and equation B

x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

x^{2} +x-5=0  

so

a=1\\b=1\\c=-5

substitute in the formula

x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}

x=\frac{-1\pm\sqrt{21}} {2}

so

The solutions are

x_1=\frac{-1+\sqrt{21}} {2}

x_2=\frac{-1-\sqrt{21}} {2}

<em>Find the values of y</em>

<em>First solution</em>

For x_1=\frac{-1+\sqrt{21}} {2}

y=2(\frac{-1+\sqrt{21}} {2})+5

y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}

The first solution is the point (\frac{-1+\sqrt{21}} {2},4+\sqrt{21})

<em>Second solution</em>

For x_2=\frac{-1-\sqrt{21}} {2}

y=2(\frac{-1-\sqrt{21}} {2})+5

y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}

The second solution is the point (\frac{-1-\sqrt{21}} {2},4-\sqrt{21})

Round to the nearest hundredth

<em>First solution </em>

(\frac{-1+\sqrt{21}} {2},4+\sqrt{21}) -----> (1.79,8.58)

(\frac{-1-\sqrt{21}} {2},4-\sqrt{21}) -----> (-2.79,-0.58)

see the attached figure to better understand the problem

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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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