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geniusboy [140]
3 years ago
9

I was wondering how you do the equation f (x)=6x^3+8?

Mathematics
1 answer:
Rus_ich [418]3 years ago
6 0
0=6x^3+8
-8           -8
-8=6x^3
----   ------
  6       6
(-8)/6=x^3
cube root each side to get
∛[(-8)/6]=x or F(x)=∛[(-8)/6]
F(x)=-1.100642...
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Two lines, A and B, are represented by equations given below: Line A: y = x − 4 Line B: y = 3x + 4 Which of the following shows
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What is the median number of guest for each holiday
Ivan

You have not given the data for which you like to find the median.

I will however explain how you can find the median of a given set of data, and you can apply the same method to your problem.

Step-by-step explanation:

Median, just like it sounds, is simply the middle value of a given set of data.

Given a set of numbers:

a, b, c, d, e, ..., z.

To find the median, first,

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If it is even, you have two middle numbers, then the median will be the addition of those two numbers divided by two.

Example: To find the median of

4, 5, 2, 1, 2, 6, 7, 3, 5.

First, we arrange in ascending:

1, 2, 2, 3, 4, 5, 5, 6, 7

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The MEDIAN is 4 .

Example 2: To find the median of

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3 0
3 years ago
Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra
bearhunter [10]

Answer:

\frac{dh}{dt}=-\frac{1}{2\pi}cm/min

Step-by-step explanation:

From similar triangles, see diagram in attachment

\frac{r}{4}=\frac{h}{16}


We solve for r to obtain,


r=\frac{h}{4}


The formula for calculating the volume of a cone is

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We substitute the value of r=\frac{h}{4} to obtain,


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We now differentiate both sides with respect to t to get,

\frac{dV}{dt}=\frac{\pi}{16}h^2 \frac{dh}{dt}


We were given that water is drained out of the tank at a rate of 2cm^3/min


This implies that \frac{dV}{dt}=-2cm^3/min.


Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means h=8cm.


We substitute this values to obtain,


-2=\frac{\pi}{16}(8)^2 \frac{dh}{dt}


\Rightarrow -2=4\pi \frac{dh}{dt}


\Rightarrow -1=2\pi \frac{dh}{dt}


\frac{dh}{dt}=-\frac{1}{2\pi}






3 0
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