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3 years ago

Which function is a quadratic function

1 answer:

5 0

A quadratic function is one of the form f(x) = ax2 + bx + c, where a, b, and c are numbers with a not equal to zero. The graph of a quadratic function is a curve called a parabola. ... A parabola intersects its axis of symmetry at a point called the vertex of the parabola. You know that two points determine a line.

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When comparing the graphs of y = x and y > x, what is a difference in the graphs?

mars1129 [50]

I donnt think that there is a solution, but could be wrong

8 0

3 years ago

A box is to be constructed with a rectangular base and a height of h cm. The length, l, of the box is 10 cm, and the width, w, i

Svetradugi [14.3K]

<span>The formula for the volume of a box is: Volume = L*W*H We know that the length is 10 cm. We know that the width is twice the height, so W = 2H. Then our volume equation is: Volume = 10*2H*H Volume = 20*H^2 Since H^2 is in the equation, this is a quadratic equation.</span>

8 0

3 years ago

The diameter and height of this cylinder are equal to the side length,S,of the cube in which the cylinder is imscribed. What is

max2010maxim [7]

$V=B\times H$
$V=\pi r^{2}h$
$V=\pi (\frac{d}{2})^{2}h$
$V=\pi (\frac{s}{2})^{2}s$
$V=\pi (\frac{s^{2}}{4})s$
$V=\frac{\pi s^{3}}{4}$

3 0

4 years ago

50 POINTS FOR THIS

sergejj [24]

Answer:

i have the same problem as you, i need help as well

Step-by-step explanation:

6 0

3 years ago

Question 5 and 6 please help me

sp2606 [1]

Problem 5

The function is continuous for the given domain $x \ge 6$

This is because y = (-5/6)x+5 is itself continuous, and any interval subset of this function is also continuous. We can plug in any real number that is equal to 6 or larger, and get some y output. If we plugged in x = 6, then we'd get

y = (-5/6)x+5

y = (-5/6)*6 + 5

y = -5+5

y = 0

This is the largest y value possible. Why? Because y = (-5/6)x+5 has a negative slope, so the graph is going downhill as you read it from left to right. As x gets bigger, y gets smaller. The smallest x value allowed in the domain produces the largest y value in the range. There is no smallest y value as the y values keep going down forever.

The range is therefore $y \le 0$

In interval notation, you can write the range as $(-\infty, 0]$ . The square bracket indicates "include this endpoint as part of the range".

======================================================

Problem 6

The function is discrete for this given domain. The domain itself is a discrete list of values. We cannot plug in values between say 0 and 2. We can only substitute one of those values from the list given. Consequently, the y values will also be a list, and not an interval like problem 5 had.

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If you plugged in x = -4, then you should get...

y = (-1/2)*(-4)+2

y = 2+2

y = 4

So the input x = -4 lead the output y = 4

Repeat for x = -2

y = (-1/2)x+2

y = (-1/2)*(-2)+2

y = 1+2

y = 3

and the same for x = 0 as well

y = (-1/2)x+2

y = (-1/2)*0 + 2

y = 0 + 2

y = 2

and x = 2 also

y = (-1/2)x+2

y = (-1/2)*2 + 2

y = -1+2

y = 1

Finally, plug in x = 4

y = (-1/2)x+2

y = (-1/2)*4+2

y = -2+2

y = 0

---------------

If we plugged each of these x values {-4, -2, 0, 2, 4} one at a time into the equation y = (-1/2)x+2, then we get this list of values {4, 3, 2, 1, 0}

Sorting the values from smallest to largest, we have this range {0, 1, 2, 3, 4}

3 0

3 years ago