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2 years ago

Question 21..........

Mathematics

1 answer:

Rudiy272 years ago

5 0

When the exponent is a negative value, you would move the decimal point that many places to the left.

9 x 10^-4 = 0.0009

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What is the inverse of the function f(x) = x +3?

OLga [1]

Answer:

h(x) = x - 3

Step-by-step explanation:

To find the inverse of a function, you can switch the x and y variables.

In f(x) = x + 3, the y variables is "f(x)".

y = x + 3

Now switch the x and y.

x = y + 3

Isolate for y and replace with h(x).

x - 3 = y

y = x - 3

h(x) = x - 3

5 0

2 years ago

Read 2 more answers

Complete the sentences below:

pogonyaev

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3 0

3 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago

The numbers of regular season wins for 10 football teams in a given season are given below. Determine the range, mean, variance,

tensa zangetsu [6.8K]

We have the following data set:

2,7,15,3,12,9,15,8,3,10

The range is the difference between the highest and lowest values in the set, to find the range, order the data set from least to greatest.

2,3,3,7,8,9,10,12,15,15

Then,

$\begin{gathered} \text{Range}=15-2 \\ \text{Range}=13 \end{gathered}$

Mean is represented by the following expression:

$\text{Mean}=\frac{\text{Sum of all data points}}{Number\text{ of data po}ints}$ $\text{Mean}=\frac{84}{10}=8.4$

Population variance formula looks like this:

$\begin{gathered} \sigma^2=\frac{\sum^{}_{}(x-\mu)^2}{N} \\ \text{where,} \\ \sigma^2=\text{population variance} \\ \sum ^{}_{}=addition\text{ of} \\ x=\text{each value} \\ \mu=population\text{ mean} \\ N=\text{ number of values in the population} \end{gathered}$

Then, substituting: