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2 years ago

Find the larger of the two roots of the equation y = -2.2x2 + 63.5x - 17

Mathematics

1 answer:

Luden [163]2 years ago

7 0

The larger of the two roots of the equation $y = -2.2x^2 + 63.5x - 17$ would be 28.59.

<h3>How to find the roots of a quadratic equation?</h3>

Suppose that the given quadratic equation is

$ax^2 + bx + c = 0$

Then its roots are given as:

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

The given quadratic equation is

$y = -2.2x^2 + 63.5x - 17$

now, the roots of the equation are

$x = \dfrac{-63.5 \pm \sqrt{63.5^2 - 4(-2.2)(-17)}}{2(-2.2)}\\\\\\x = \dfrac{-63.5 \pm \sqrt{4032.25 - 149.6}}{(-4.4)}\\\\\\x = \dfrac{-63.5 \pm \sqrt{3882.65}}{(-4.4)}\\\\\\x = \dfrac{-63.5 \pm 62.31}{(-4.4)}$

The two roots are 0.27 and 28.59.

Thus, the larger of the two roots of the equation $y = -2.2x^2 + 63.5x - 17$ would be 28.59.

Learn more about finding the solutions of a quadratic equation here:

brainly.com/question/3358603

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Determine which function has the greatest rate of change over the interval [0, 2].

ruslelena [56]

Remember that the average rate of change of a function over an interval is the slope of the straight line connecting the end points of the interval. To find those slopes, we are going to use the slope formula: $m= \frac{y_{2}-y_{1}}{x_2-x_1}$

Rate of change of $a$ :
From the graph we can infer that the end points are (0,1) and (2,4). So lets use our slope formula to find the rate of change of $a$ :
$m= \frac{y_{2}-y_{1}}{x_2-x_1}$
$m= \frac{4-1}{2-0}$
$m= \frac{3}{2}$
$m=1.5$
The average rate of change of the function $a$ over the interval [0,2] is 1.5

Rate of change of $b$ :
Here the end points are (0,0) and (2,2)
$m= \frac{2-0}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $b$ over the interval [0,2] is 1

Rate of change of $c$ :
Here the end points are (0,-1) and (2,0)
$m= \frac{0-(-1)}{2-0}$
$m= \frac{1}{2}$
$m=0.5$
The average rate of change of the function $c$ over the interval [0,2] is 0.5

Rate of change of $d$ :
Here the end points are (0,0.5) and (2,2.5)
$m= \frac{2.5-0.5}{2-0}$
$m= \frac{2}{2}$
$m=1$
The average rate of change of the function $d$ over the interval [0,2] is 1

We can conclude that the <span>function that has the greatest rate of change over the interval [0, 2] is the function a.</span>

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Answer:

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Step-by-step explanation:

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