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3 years ago

Find the number of terms and the degree of this polynomial.

Mathematics

1 answer:

dangina [55]3 years ago

5 0

Answer:

Number of terms:3

Degree:2nd degree

Step-by-step explanation:

-5c^2 +8c +2-3

-5c^2 +8c-1

he 2nd degree is the power of 2

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Amare wants to ride a Ferris wheel that sits four meters above and has a diameter of 50 meters. It takes 6 minutes to do three r

Veseljchak [2.6K]

To answer this question, we begin to transform the speed of revolutions per minute in radians per minute.
They tell us that the speed of the wheel is 3 turns in 6 minutes.
So:
1 revolution = 2π
(2π / 1 turn) * (3 turns / 6min) = 6π/ 6min = 1π / min.
The wheel turns π or 180 ° in one minute.
We already have the angular velocity w.
We know that the lowest point of the wheel is 4 meters above the ground and that it returns to the same point every 2π
Therefore, the function sought is periodic and must be equal to 4 for allvalues of time k, where k is an even number 2, 4, 6, 8, .., k

Then the function must have the form rsin(wt) where "t" is the elapsed time, "w" is the previously calculated angular velocity, and "r" is the radius of the wheel.
The minimum value of the function must be 4 and the maximum value 54.
Therefore, the function is:

h (t) = 4 + 25 + 25sin (π×t + 3π / 2)
Where 3π/ 2 is the phase angle, which indicates that the movement starts at the instant t = 0 at the lowest point of the wheel that equals 3π / 2.
You can verify the answer in the following way:

After 1 minute, the wheel should have rotated 180 ° or π. Therefore, the person must be at the highest point of the wheel and his height must be 54 m.
When you replace t = 1 in the formula, you get h = 54m
After 2 minutes, the wheel should have rotated 360 ° or 2π. Therefore, the person must be at the starting point and their height must be 4 m.
By replacing t = 2 in the formula you will get h = 4m
After 0.5 minutes, the wheel should have rotated 90 ° or π / 2. Therefore, the person must be in the right half of the wheel and his height must be 29 m.
When replacing t = 0.5 in the formula you will get h = 29m

4 0

4 years ago

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

Line AB is rotated to form A'B'¯¯¯¯¯¯¯ .

sleet_krkn [62]

I think the correct answer is A. Hope this helps.

6 0

3 years ago

Read 2 more answers

PLEASE ANSWER QUICKLY

Strike441 [17]

ANSWERS AND STEPS???

3 0