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3 years ago

What is 1/3 of a kilometer

Mathematics

1 answer:

marissa [1.9K]3 years ago

7 0

= (1/3) * 1000 meters
= 333.33 meters
OR
=.3333333 kilometers

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Disuss the difference between fixed expenses and variable expenses as they relate to a budget?

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4 years ago

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3 years ago

Let f(x)=3x-6. Is f one-to-one? If it is find f^-1, if not explain why.

beks73 [17]

Answer: Yes, $\bold{f^{-1}(x)=\dfrac{x+6}{3}}$

<u>Step-by-step explanation:</u>

f(x) = 3x - 6 is a line so it is a function because

it passes the vertical line test

and the horizontal line test

To find the inverse, swap the x's and y's and solve for y

$y=3x-6\\\\\\\text{Swap the x's and y's:}\qquad x=3y-6\\\\\text{Add 6 to both sides:}\qquad x+6=3y\\\\\text{Divide both sides by 3:}\qquad \dfrac{x+6}{3}=y$

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3 years ago

We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure

STALIN [3.7K]

Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be $h(t) = -16\cdot t^{2} + 128\cdot t + 320$ , the first and second derivatives are, respectively:

First Derivative

$h'(t) = -32\cdot t +128$

Second Derivative

$h''(t) = -32$

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

$-32\cdot t +128 = 0$

$t = \frac{128}{32}\,s$

$t = 4\,s$ (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

$h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320$