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3 years ago

Suppose that Y and Z are points on the number line.

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

6 0

Answer:

16-17 is the answer

although I am not sure

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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 $\vec{v} = 2i+2j$

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,

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

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

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2 years ago

Whats 9*20(10+20) First Answer Gets Brainliest

Tcecarenko [31]

Answer:

Your answer would be 5400.

Step-by-step explanation:

You have to follow PEMDAS for this one. You will start within the parenthesis and add 10 and 20 together to get 30. You would then times that by 20 and get 600. Finally, you would mulitply that by 9 and get 5400 as your answer..

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3 years ago

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Elena L [17]

Answer: I helped LOL

Step-by-step explanation:

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devlian [24]

Answer:

The y intercept is 0 and 3

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Triss [41]

Answer:I think it is a

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