Find the distance from E to F

Mathematics

1 answer:

ludmilkaskok [199]3 years ago

7 0

Answer/Step-by-step explanation:

To find the distance of E to F, subtract the distance of DE from the total.

28 - 11 = 17

EF = 17

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Rounded to the nearest hundredth, what is the positive solution to the quadratic equation 0 = 2x2 + 3x – 8? Quadratic formula: x

Lynna [10]

ANSWER

1.39

EXPLANATION

The given quadratic equation is

$0 = 2 {x}^{2} + 3x - 8$

This is the same as,

$2 {x}^{2} + 3x - 8 = 0$

Comparing to

$a {x}^{2} + bx + c = 0$

We have

a=2, b=3,c=-8

Using the quadratic formula, the solution is given by:

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

We substitute the values to get,

$x = \frac{ - 3\pm \: \sqrt{ {3}^{2} - 4(2)( - 8)} }{2(2)}$

$x = \frac{ - 3\pm \: \sqrt{ 73} }{4}$

The positive root is

$x = \frac{ - 3 + \: \sqrt{ 73} }{4} = 1.39$

to the nearest hundredth.

4 0

3 years ago

Read 2 more answers

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .