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3 years ago

8

What expression is equivalent to 4(2x+3)

1 answer:

yuradex [85]3 years ago

4 0

Answer: 8x+12

Distribute the 4. So, 4(2x)+4(3). Then you end up with 8x+12.

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Evaluate the sum from n equals five to nine of quantity three n plus two

goldfiish [28.3K]

Answer:

The given expanded sum of the series is $\sum\limits_{n=5}^{9}3n+2=115$

Step-by-step explanation:

Given problem can be written as

$\sum\limits_{n=5}^{9}3n+2$

To find their sums:

Now expanding the series

That is put n=5,6,7,8,9 in the given summation

$\sum\limits_{n=5}^{9}3n+2=[3(5)+2]+[3(6)+2]+[3(7)+2]+[3(8)+2]+[3(9)+2]$

$=[15+2]+[18+2]+[21+2]+[24+2]+[27+2]$

$=17+20+23+26+29$ (adding the terms)

$=115$

Therefore $\sum\limits_{n=5}^{9}3n+2=115$

Therefore the given sum of the series is $\sum\limits_{n=5}^{9}3n+2=115$

The given expanded sum of the series is $\sum\limits_{n=5}^{9}3n+2=115$

6 0

3 years ago

2 1/3 divided by 1/4

mario62 [17]

Answer:

9 1/3 or 28/3

Step-by-step explanation:

3 0

3 years ago

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Write an equation for the line, given that M=5/6 and why intercept is (0,12)

Fudgin [204]

The answer to the question

6 0

3 years ago

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Need help on 4,5 pleasesss

Tamiku [17]

The answer is 45 you put the 4 and the 5 togther

7 0

3 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