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3 years ago

Does the sum of -21 and 12 equal 0?

Mathematics

1 answer:

Orlov [11]3 years ago

6 0

No the sum is -21 is bigger than 12 because

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makvit [3.9K]

Answer:

900

Step-by-step explanation:

2pi = 360

1pi = 180

4 0

3 years ago

3/12 of an orange how many 4ths do you have

Anika [276]

1/4 its like dividing

7 0

4 years ago

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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

Elijah earns $200 for 8 hours what is the unit rate per hour

Fiesta28 [93]

Answer:

$25 per hour

Step-by-step explanation:

You would divide 200 by 8 and you'll get 25

6 0

4 years ago

Each shape is 1 whole. Which fraction greater than 1 names the parts that are shaded

Nutka1998 [239]

You have no pictures of the shapes?

6 0

3 years ago

Read 2 more answers