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3 years ago

What are the coordinates of the terminal point determined by t= 11 pi / 3

Mathematics

2 answers:

Naddika [18.5K]3 years ago

6 0

Answer: B

Step-by-step explanation: Hope it helps

LuckyWell [14K]3 years ago

5 0

Answer:

hope this works!

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Find the p-value for the indicated hypothesis test with the given standardized test statistic, z. decide whether to reject upp

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3 years ago

The cost of a pen at the office supply store is $2.65. The school store sells the pen for $3.45. How much does the school store

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3 years ago

Evaluate the interval (Calculus 2)

Darya [45]

Answer:

$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

3 0

1 year ago

The equation x2 - 3x = 22

gizmo_the_mogwai [7]

Answer:

x = - 3.42443 or 6.42443

Step-by-step explanation:

5 0

3 years ago

The coordinates of the vertices of ΔPQR are (–2, –2), (–6, –2), and (–6, –5).

melamori03 [73]

<h3>Answer:</h3>

Yes, ΔPʹQʹRʹ is a reflection of ΔPQR over the x-axis

<h3>Explanation:</h3>

The problem statement tells you the transformation is ...

... (x, y) → (x, -y)

Consider the two points (0, 1) and (0, -1). These points are chosen for your consideration because their y-coordinates have opposite signs—just like the points of the transformation above. They are equidistant from the x-axis, one above, and one below. Each is a <em>reflection</em> of the other across the x-axis.

Along with translation and rotation, <em>reflection</em> is a transformation that <em>does not change any distance or angle measures</em>. (That is why these transformations are all called "rigid" transformations: the size and shape of the transformed object do not change.)

An object that has the same length and angle measures before and after transformation <em>is congruent</em> to its transformed self.

So, ... ∆P'Q'R' is a reflection of ∆PQR over the x-axis, and is congruent to ∆PQR.

6 0

3 years ago