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3 years ago

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Identify the maximum and minimum values of the function y = 3 cos x in the interval [–2π, 2π]. Use your understanding of transfo

rmations, not your graphing calculator.

1 answer:

8_murik_8 [283]3 years ago

5 0

You would need to do this with a graphing calculator

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Graph the image of triangle PQR after a reflection across the line y = 2

ololo11 [35]

Answer: Refer to the diagram below

P ' is at (-7, 9)
Q ' is at (1, 9)
R ' is at (-6, 4)

================================================

Explanation:

The diagram shows how point R moves to R'. We move up 2 units going from R(-6,0) to (-6,2). This lands us on the line of reflection. Then we move another 2 units up to land on (-6,4) which is the location of point R'.

The other points P and Q follow the same idea. Though the distances will be different from R. For P and Q, we'll move 7 units up to arrive at the line of reflection, then another 7 units to arrive at the proper locations of P' and Q', which are (-7,9) and (1,9) respectively.

3 0

3 years ago

Somebody help me with this?

trasher [3.6K]

Answer:

Zero's

Step-by-step explanation:

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2 years ago

What is the absolute deviation of 4 in this data set?

Nana76 [90]

Answer:

n

6

Mean

11

Average Absolute Deviation (MAD)

4.6667

Step-by-step explanation:

4 0

2 years ago

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions

Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector $\vec{v}$ in some vector space $\mathbb R^n$ we have

$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

$\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}$

is preferable because the inner product on $\mathbb R^n$ is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as $\vec{v} = 2i+2j$

4.

For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

$\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}$

Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

3 0

2 years ago

The formula for the cost of buying a car=

svet-max [94.6K]

8000-2600=5400
5400/12=450
monthly pavement =450
12x450+2600=8000

5 0

3 years ago