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MrRa [10]
3 years ago
14

Which statement is true about the extreme value of the given quadratic equation? A. The equation has a maximum value with a y-co

ordinate of -21. B. The equation has a maximum value with a y-coordinate of -27. C. The equation has a minimum value with a y-coordinate of -21. D. The equation has a minimum value with a y-coordinate of -27.
Mathematics
1 answer:
insens350 [35]3 years ago
3 0

Answer:

A. The equation has a maximum value with a y-coordinate of -21.

Step-by-step explanation:

From the given equation:

\mathbf{y = -3x^2 + 12x -33}

This parabola is vertical and is goes downward via the negative path

Where the vertex represents the maximum value;

\mathbf{y = -3 (x^2 + 4x) -33}

Using completing the square method;

\mathbf{y = -3 (x^2 + 4x+2^2) -33+12}

\mathbf{y = -3 (x^2 + 4x+4) -21}

To perfect square:

\mathbf{y = -3 (x-2)^2 -21}

The vertex point is (2, -21)

Hence ; the equation has a maximum value with a y-coordinate of -21.

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stiv31 [10]
Answer
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First we write out what we know
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3n + 2f = 30

Because we know from the first equation
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3n + 2(n+5) = 30
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5n +10 =30. Now isolate n by subtracting 10 from both sides
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Now we do the same thing to find f
We substitute the value of n (4) into the equation 3n + 2f =30
3(4) +2f =30
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2f = 18. Now isolate f by dividing both sides by 2
f = 9


We check our work by inserting the n and f values we found into one of the equations
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1 year ago
Solve for x.<br> y = (5+x}m<br> Answer?
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Answer:

x=ym−4

Step-by-step explanation:

y=(4+x)m

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(I really hope this helps)

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3 years ago
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Answer:

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cestrela7 [59]

Answer:

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