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Verizon [17]
3 years ago
5

What is 1/s as a single term without a denominator?

Mathematics
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

consistent

Step-by-step explanation:

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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
Drag and drop the correct surface area to match the cube. cube with edge length = 3.2 m 12.8 m²19.2 m²51.2 m²61.44 m²
Ira Lisetskai [31]
- Surface Area of a Cube = 6(L)²
- Surface Area of this cube = 6(3.2)²
                                              = 61.44 m²
3 0
3 years ago
Can someone please help me​
Svetllana [295]

Step-by-step explanation:

I think its 2/6 becuase its bigger

3 0
3 years ago
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liq [111]

Answer:

6 units

Step-by-step explanation:

count 6 units up

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Answer:

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underfined

Step-by-step explanation:

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