Find lim (x2-4)/(x+2), Evaluate integral from 0 to 1 of (3x2 + 2x + 4) dx, Find dy/dx for y = (2x3 - 5)3…

Question

Find lim (x2-4)/(x+2), Evaluate integral from 0 to 1 of (3x2 + 2x + 4) dx, Find dy/dx for y = (2x3 - 5)3, Sin100 Cos 80 + Sin80 cos100 is, Convert 5pi/3 into degree, The angle 210 lies on :, Find the interquartile range for the data 4, 6, 9, 10, 17, 19, and 20

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Here are the solutions to the questions:

1. Find $\lim_{x \to -2} \left(\frac{x^2-4}{x+2}\right)$

Factor the numerator.
The numerator $x^2-4$ is a difference of squares, which can be factored as $(x-2)(x+2)$ .
$\frac{x^2-4}{x+2} = \frac{(x-2)(x+2)}{x+2}$
For $x \neq -2$ , we can cancel out the $(x+2)$ term.
$\frac{(x-2)(x+2)}{x+2} = x-2$
Evaluate the limit by substituting $x = -2$ .
$\lim_{x \to -2} (x-2) = -2 - 2 = -4$
The limit is $\boxed{-4}$ .

The correct option is B.

2. Evaluate $\int_{0}^{1} (3x^2 + 2x + 4)\,dx$
Integrate each term of the polynomial.
$\int (3x^2 + 2x + 4)\,dx = \frac{3x^{2+1}}{2+1} + \frac{2x^{1+1}}{1+1} + 4x + C$ $= \frac{3x^3}{3} + \frac{2x^2}{2} + 4x + C$ $= x^3 + x^2 + 4x + C$
Evaluate the definite integral using the limits of integration.
$\left[x^3 + x^2 + 4x\right]_{0}^{1} = (1^3 + 1^2 + 4(1)) - (0^3 + 0^2 + 4(0))$ $= (1 + 1 + 4) - (0)$ $= 6$
The value of the integral is $\boxed{6}$ .

The correct option is C.

3. Find $\frac{dy}{dx}$ for $y = (2x^3 - 5)^3$
Apply the chain rule.
Let $u = 2x^3 - 5$ . Then $y = u^3$ .
First, find the derivative of $y$ with respect to $u$ :
$\frac{dy}{du} = \frac{d}{du}(u^3) = 3u^2$
Next, find the derivative of $u$ with respect to $x$ :
$\frac{du}{dx} = \frac{d}{dx}(2x^3 - 5) = 2(3x^{3-1}) - 0 = 6x^2$
Multiply the derivatives to find $\frac{dy}{dx}$ .
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (3u^2)(6x^2)$
Substitute $u = 2x^3 - 5$ back into the expression:
$\frac{dy}{dx} = 3(2x^3 - 5)^2 (6x^2)$ $\frac{dy}{dx} = 18x^2 (2x^3 - 5)^2$
The derivative is $\boxed{18x^2 (2x^3 - 5)^2}$ .

None of the provided options match this result. There might be a typo in the question or the options.

4. $\sin 10^\circ \cos 80^\circ + \sin 80^\circ \cos 10^\circ$ is
Recognize the sum formula for sine.
The formula for $\sin(A+B)$ is $\sin A \cos B + \cos A \sin B$ .
In this expression, $A = 10^\circ$ and $B = 80^\circ$ .
So, the expression can be written as $\sin(10^\circ + 80^\circ)$ .
Calculate the sum of the angles and evaluate the sine.
$\sin(10^\circ + 80^\circ) = \sin(90^\circ)$
We know that $\sin(90^\circ) = 1$ .
The value of the expression is $\boxed{1}$ .

The correct option is A.

5. Convert $\frac{5\pi}{3}$ into degree
Use the conversion factor from radians to degrees.
To convert radians to degrees, multiply by $\frac{180^\circ}{\pi}$ .
$\frac{5\pi}{3} \text{ radians} \times \frac{180^\circ}{\pi \text{ radians}}$
Perform the calculation.
$\frac{5\pi}{3} \times \frac{180^\circ}{\pi} = \frac{5 \times 180^\circ}{3}$ $= 5 \times 60^\circ$ $= 300^\circ$
The angle in degrees is $\boxed{300^\circ}$ .

The correct option is B.

6. The angle $210^\circ$ lies on :
Recall the ranges for each quadrant.
- First quadrant: $0^\circ < \theta < 90^\circ$
- Second quadrant: $90^\circ < \theta < 180^\circ$
- Third quadrant: $180^\circ < \theta < 270^\circ$
- Fourth quadrant: $270^\circ < \theta < 360^\circ$
Determine where $210^\circ$ falls within these ranges.
Since $180^\circ < 210^\circ < 270^\circ$ , the angle $210^\circ$ lies in the third quadrant.
The angle $210^\circ$ lies on the $\boxed{\text{third quadrant}}$ .

The correct option is C.

7. Find the interquartile range for the data 4, 6, 9, 10, 17, 19, and 20
Order the data.
The data is already in ascending order: 4, 6, 9, 10, 17, 19, 20.
There are $n=7$ data points.
Find the median ( $Q_2$ ).
The median is the middle value. For $n=7$ , the median is the $\frac{7+1}{2} = 4^{th}$ value.
$Q_2 = 10$ .
Find the first quartile ( $Q_1$ ).
$Q_1$ is the median of the lower half of the data (excluding $Q_2$ for an odd number of data points).
Lower half: 4, 6, 9.
The median of these three values is the $\frac{3+1}{2} = 2^{nd}$ value.
$Q_1 = 6$ .
Find the third quartile ( $Q_3$ ).
$Q_3$ is the median of the upper half of the data (excluding $Q_2$ ).
Upper half: 17, 19, 20.
The median of these three values is the $\frac{3+1}{2} = 2^{nd}$ value.
$Q_3 = 19$ .
Calculate the interquartile range (IQR).
The interquartile range is $IQR = Q_3 - Q_1$ .
$IQR = 19 - 6 = 13$
The interquartile range is $\boxed{13}$ .

The correct option is D.

Find lim (x2-4)/(x+2), Evaluate integral from 0 to 1 of (3x2 + 2x + 4) dx, Find dy/dx for y = (2x3 − 5)3, Sin100 Cos 80 + Sin80 cos100 is, Convert 5pi/3 into degree, The angle 210 lies on :, Find the interquartile range for the data 4, 6, 9, 10, 17, 19, and 20

Answer

Step-by-step solution

Factor the numerator.

Evaluate the limit by substituting $x = -2$ .

Integrate each term of the polynomial.

Evaluate the definite integral using the limits of integration.

Apply the chain rule.

Multiply the derivatives to find $\frac{dy}{dx}$ .

Recognize the sum formula for sine.

Calculate the sum of the angles and evaluate the sine.

Use the conversion factor from radians to degrees.

Perform the calculation.

Recall the ranges for each quadrant.

Determine where $210^\circ$ falls within these ranges.

Order the data.

Find the median ( $Q_2$ ).

Find the first quartile ( $Q_1$ ).

Find the third quartile ( $Q_3$ ).

Calculate the interquartile range (IQR).

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