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3 years ago

HELP !!! Evaluate. 0^3 =

Mathematics

2 answers:

egoroff_w [7]3 years ago

6 0

Anything multiplied by zero is zero, even if you do it several times.

Diano4ka-milaya [45]3 years ago

4 0

Think of it as 0 times 0 times 0

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Determine the value of x if (211) base x = 152 base 8

tekilochka [14]

It's the same as x^211 = 8^152

Simplify that equation. Find the 211th root of (8^152), and that is what x equals. If I remember logs correctly...

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4 years ago

I need the answer pls

Pie

Answer:

Step-by-step explanation:

Subtract-4 from 6

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3 years ago

What is the place value relationship of the 9's in 299

eimsori [14]

Ones and tens thats the anwers

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3 years ago

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A) 289.6 divide by 6.4 b) 236.592 divide by 0.36

Simora [160]

Answer:

a)45.25

b)657.2

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3 years ago

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Please help very urgent

Alik [6]

Answer:

32, 16, 8, 4, 2, 1

Explanation:

The geometric mean can be represented by $\sqrt[n]{x_{1} • x_{2} • x_{3} • .. x_{n}}$ .

Which is the mean of the product of n numbers, used to find the average of a geometric progression.

Don't get confused by geometric mean, it is only asking you about the next numbers in the geometric sequence given the first and sixth term.

The explicit rule for a geometric sequence can be modeled by:

$a_{n} = a_{1} • r^{n-1}$

Where $a_{n}$ is the nth term, $a_{1}$ is the first term in the sequence, n is the term number, and r is the common ratio.

Since we already know the first term, $a_{1}$ will simply be 32.

Since we know it's geometric, there will be an exponential relationship, which means that we will use the geometric mean to find the common ratio.

There are 6 total terms, r is raised to the n – 1 so 6 – 1 = 5, and that will be the degree of this root.

$\sqrt[5]{\frac{a_{6}}{a_{1}}}$ =

$\sqrt[5]{\frac{1}{32}}$ =

$\frac{1}{2}$ .

Therefore: $r = \frac{1}{2}$ .

Using all the information we have, we can find the explicit rule:

$a_{n} = a_{1} • r^{n-1}$

$a_{1}$ = 32
$r = \frac{1}{2}$

$a_{n} = a_{1} • r^{n-1}$ →

$\boxed{a_{n} = 32 • (\frac{1}{2})^{n-1}}$

________________________________

We can test that this works by substituting the number location of the term you want to find.

For instance:

$a_{1} = 32 • (\frac{1}{2})^{1-1}$

$a_{1} = 32 • (\frac{1}{2})^{0}$

$a_{1} = 32 • 1$

$a_{1} = 32$

$a_{6} = 32 • (\frac{1}{2})^{6-1}$

$a_{6} = 32 • (\frac{1}{2})^{5}$

$a_{6} = 32 • \frac{1}{32}$

$a_{6} = 1$

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3 years ago

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