What does the magnitude of a vector represent?

Mathematics

1 answer:

nalin [4]3 years ago

7 0

Answer:

A vector's magnitude represents its length, so your answer is C, the length of a vector.

Step-by-step explanation:

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Can somebody help me? I’m already failing :(

Yuliya22 [10]

Answer:

125

Step-by-step explanation:

f(1)= the first time (32,000)

n= 9

r= .5 (each term is being mulitplied by .5 to get the next one

We have

32,000*(.5)^(9-1)

32,000*(.5)^8

=125

5 0

2 years ago

What is the distance between point A and point B?what is the ordered pair for each of the points plotted above?

Alekssandra [29.7K]

Answer:

A: (0,0), B: (3, -4), 5 units

Step-by-step explanation:

A: (0,0), B: (3, -4)

To find the distance, use the distance formula. Square-root((0-3)^2+(0-(-4))^2) ---> Square-root((-3)^2 + 4^2) ---> Square-root(9+16) ---> Square-root(25) ---> 5units.

6 0

3 years ago

HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$