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scoray [572]
3 years ago
11

2. Cory makes a map of his favorite park, using a coordinate system with units of yards. The old oak tree is at position (1, 10)

and the granite boulder is at position (-5, 9). How far apart are the old oak tree and the granite boulder? Round to the nearest tenth if necessary
Mathematics
1 answer:
Alex17521 [72]3 years ago
7 0

Answer: The old oak tree and the granite boulder are \sqrt{26} units apart.

Step-by-step explanation:

The distance between points (a,b) and (c,d) is given by :-

d=\sqrt{(c-a)^2+(d-b)^2}

Given: The position of oak tree = (1, 10)

The position of granite boulder = (-5,9)

The distance between oak tree and granite boulder  = \sqrt{(10-9)^2+(1-(-5))^2}

=\sqrt{(1)^2+(1+5)^2}\\\\=\sqrt{1+25}\\\\=\sqrt{26}\text{ units}

Hence, the old oak tree and the granite boulder are \sqrt{26} units apart.

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Morten Andersen played in NFL for 25 years write and solve an equation to find how many points he averaged each year
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Answer:

r =\frac{2437points}{25 years} =97.45 \frac{points}{year}

So then we have that Morten Andersen scored on average 97.45 points per year in his career.

Step-by-step explanation:

Assuming the following table on the figure attached.

We see that the career points for Morten Andersen was 2437. That include all the points over alll the years the he played in the NFL.

Since the total years played by Morten Andersen was 25 we can write the following equation:

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Where r represent the rate of points average per year.

If we solve for r from the last equation we can divide both sides of the equation and we got:

r =\frac{2437points}{25 years} =97.45 \frac{points}{year}

So then we have that Morten Andersen scored on average 97.45 points per year in his career.

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Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x
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Answer:

(a) The function is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is x = -\frac{1}{2}

(d) The function is concave upward on \left(- \frac{1}{2}, \infty\right) and concave downward on \left(-\infty, - \frac{1}{2}\right)

Step-by-step explanation:

(a) To find the intervals where f(x) = 2x^3 + 3x^2 -180x is increasing or decreasing you must:

1. Differentiate the function

\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6

These points divide the number line into three intervals:

(-\infty,-6), (-6,5), and (5, \infty)

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right

Therefore f(x) is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

  • f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.
  • f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6

We set f''(x) = 0

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\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right

The function is concave upward on (-1/2,\infty) because the f''(x) > 0 and concave downward on (-\infty,-1/2) because the f''(x) < 0.

f(x) is concave down before x = -\frac{1}{2}, concave up after it. So f(x) has an inflection point at x = -\frac{1}{2}.

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