All categories

1 year ago

Find the distance between (-4,3,-6) and the origin ((X,Y,Z) Graph)

Mathematics

1 answer:

devlian [24]1 year ago

4 0

The distance between (-4,3,-6) and the origin is $\sqrt{61$

<h3>How to determine the distance?</h3>

The point is given as:

(x, y, z) = (-4,3,-6)

The origin is

(x, y, z) = (0,0,0)

The distance between the point and the origin is calculated as:

$d = \sqrt{x^2 + y^2 + z^2$

So, we have:

$d = \sqrt{(-4)^2 + (3)^2 + (-6)^2$

Evaluate

$d = \sqrt{61$

Hence, the distance between (-4,3,-6) and the origin is $\sqrt{61$

You might be interested in

Find the population mean or sample mean as indicated. Sample: 19, 15, 6, 11, 24 Compute the sample mean for this data set. Selec

yKpoI14uk [10]

Answer:

$\overline{x}=15$

Step-by-step explanation:

the mean is given by:

$\overline{x} = \dfrac{\sum\limits_{i=1}^n x_i}{n} \quad\text{or}\quad \dfrac{\text{sum of all items}}{\text{number of items}}$

In our case this is:

$\overline{x} = \dfrac{19+15+6+11+24}{5} \Rightarrow \dfrac{75}{5}\\\\\overline{x} = 15\\\\$

side note: the main difference between sample mean and population mean is in the 'context'. However, the method to calculate them is the same.

By context I mean: if this the items are taken from some larger category for example: the ages of a few 'students' from a 'class'. Here 'students' are the sample from a larger set that is 'class'. The mean of the 'few students' will be called sample mean. In contrast, if we take the mean of the ages of the whole class then this is called population mean. (population mean == mean of the whole set)

In our case we aren't told exactly where these numbers come from, is this the whole set or a sample from it, the lack of context allows us to assume that the mean can either be population mean or sample mean. So we can safely use any symbol $\mu$ or $\overline{x}$ .

3 0

3 years ago

((5^(3))^(5))/(5^(7))

Norma-Jean [14]

Answer:

Step-by-step explanation:

equals 48 trust me ive had that question

6 0

3 years ago

Read 2 more answers

“Solve the equation. Determine if it is undefined or not.”

lozanna [386]

Answer: a) No Solution

b) Infinite Solutions (All Real Numbers)

Step-by-step explanation:

4(g + 8) = 7 + 4g

4g + 32 = 7 + 4g distributed 4 into g + 8

32 = 7 subtracted 4g from both sides

Since the statement is false because 32 ≠ 7, then there is NO SOLUTION

-4(-5h - 4) = 2(10h + 8)

20h + 16 = 20h + 16 distributed

16 = 16 subtracted 20h from both sides

Since the statement is true because 16 = 16, then there are INFINITE SOLUTIONS so x can be all real numbers.

8 0

3 years ago

NEED HELP ASAP! ILL GIVE BRAINLIEST

lana66690 [7]

The red goes into the bottom plane and the blue stays on the top

5 0

2 years ago

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Pie

Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

5 0

3 years ago