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Sveta_85 [38]
3 years ago
11

Which calculation will ALWAYS give a result greater than 1? 89 × a number less than 1 7 × a number less than 1 125 − a fraction

less than 25 37 + a fraction less than 12
Mathematics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

The answer is "Option C".

Step-by-step explanation:

Please find the correct question in the attached file.

\to 1 \frac{2}{5} = \frac{7}{5} \\\\let \\  \to \frac{7}{5} =1.4\\\\\to \frac{2}{5}= 0.4\\\\\to 1.4- 0.4\\\\ \to 1\\

Its outcome is 1.4 if we subtract from 0.4. However, a fraction of less than 2/5 may be deducted. This means that the outcome still exceeds 1, that's why choice C is correct.

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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
mote1985 [20]

Answer:

\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}

Step-by-step explanation:

To find the derivative of the function y(x)=\ln \left(\frac{x}{x^2+1}\right) you must:

Step 1. Rewrite the logarithm:

\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }

Step 2. The derivative of a sum is the sum of derivatives:

\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)

Step 3. The derivative of natural logarithm is \left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}

{\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }

Step 4. The function \ln{\left(x^{2} + 1 \right)} is the composition f\left(g\left(x\right)\right) of two functions f\left(u\right)=\ln{\left(u \right)} and u=g\left(x\right)=x^{2} + 1

Step 5.  Apply the chain rule \left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }

-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}

Return to the old variable:

- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}

The derivative of a sum is the sum of derivatives:

- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)

Step 6. Apply the power rule \frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}

\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\

\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}

Thus, \frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}

3 0
3 years ago
Answer and show work #38
Serggg [28]
Did you get your answer yet
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