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Mnenie [13.5K]
3 years ago
6

Natalie reads for 6/8 hour on 6 days each week. How many hours does Natalie read each week? Enter your answer by filling in the

boxes.
Mathematics
2 answers:
likoan [24]3 years ago
8 0

Hey!

-------------------------------------------------

Answer:

\large\boxed{Natalie~reads~4~\frac{1}{2}~hours~a~week! }

-------------------------------------------------

Solution:

~Turn six into improper fraction

6 = 6/1

~Multiply

6/8 * 6/1 = 38/8

~Simplify

38/8 = 4 1/2

-------------------------------------------------

Hope This Helped! Good Luck!

UNO [17]3 years ago
7 0

Multiply the time per day by the number of days per week.

6/8 x 6 = (6 x 6)/8 = 36/8 = 4 1/2 hours a week.

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Need help with Calculus 1 inverse trig functions
lidiya [134]

Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}

Where u is a function of x as provided:

y=3tan^{-1}(x+\sqrt{1+x^2})

If we set

u=(x+\sqrt{1+x^2})

Then

\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}

\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}

Taking the derivative of y

y'=3[tan^{-1}(x+\sqrt{1+x^2})]'

Using the change of variables

\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}

Operating

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}

\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

8 0
3 years ago
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