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3 years ago

What is the length of AC?

Mathematics

2 answers:

EleoNora [17]3 years ago

5 0

For this case, we have that by definition, the distance between two points any A and B will be the absolute value of the subtraction of their coordinates in the order that is preferred. (the distances can not be negative)

So:

We find the distance from A to a point x located at 0. ENtonces:

$| -8-0 | = | -8 | = 8$

Now we add the distance from x to C. From point x (located at zero) to point C there are 5 spaces, then:

$8 + 5 = 13$

Thus, the distance from A to C is 13

Answer:

sashaice [31]3 years ago

3 0

A is at -8 and C i5 at 5

-8 to 0 is 8 units and 0 to 5 is 5 units.

8 +5 = 13

AC is 13 units long.

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Norma-Jean [14]

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So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

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$y'=(4x^3+13x^2-12x-56)'$

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$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

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$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

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

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$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

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The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

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