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3 years ago

9

The following work represents an attempt to find the equation of the parabola which has a vertex of (0, 0) and passes through th

e point (1, 3). Is the work correct? If not, what is the error?
y=a(x-h)^2+k
0=a(0-1)^2+3
-3=a

y=a(x-h)2+k
y=-3(x)^2
y=-3x^2

A.
The equation is correct.
B.
a = –3 should have been a = 9.
C.
0=a(0-1)^2+3 should have been 3=a(1-0)^2+0
D.
-3x^2 should have been 23x^2 + 3.x

2 answers:

Anastaziya [24]3 years ago

7 0

The answer to your problem is b

larisa [96]3 years ago

3 0

The answer is (C) 0=a(0-1)^2+3 should have been3=a(1-0)^2=0

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Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

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Where $a_1=a_2=a_3=0$

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Let $b_1,b_2$ and $b_3$ such that

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$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

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$b_1=b_2$ ...(4)

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asambeis [7]

Answer:

(arranged from top to bottom)

System #3, where x=6

System #1, where x=4

System #7, where x=3

System #5, where x=2

System #2, where x=1

Step-by-step explanation:

System #1: x=4

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Plug this value of y into your first equation.

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Plug this value of y into your second equation.

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Plug your value of y into your second equation.

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