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3 years ago

10

What is 0.9 written as a decimal

2 answers:

arsen [322]3 years ago

6 0

It is already in a decimal

I think you mean percent, so it will be 90%

Sorry if that is not what you mean

ankoles [38]3 years ago

6 0

Hello from MrBillDoesMath!

Answer:

0.9

Discussion:

0.9 is a decimal value

Thank you,

MrB

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Raj gets a 1.5 mile head start and runs at a rate of 4.5 miles per hour. Jacinda's progress is represented by a graph that goes

Alexxandr [17]

From the given ordered pair representing Jacinda's progress, her speed is equal to 10 miles per hour. Letting x be the time it takes for Jacinda to catch up with Raj, the total distance Raj would travel is
1.5 + 4.5x
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1.5 + 4.5x = 10x
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3 years ago

A large pizza cost $12. Each toooo g cost an additional $1.50. This situation can be repeated by the equation. Y= $2.59x +13, wh

Kay [80]

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3 years ago

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Over [174]

Answer:

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=40.75

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3 years ago

You can walk 3/4 of a mile in 1/6 of an hour (10 minutes). How far can you walk in an hour?

Hoochie [10]

Answer:

4 and 1/2 miles (or 9/2 miles)

Step-by-step explanation:

I would set this up as a proportion

3/4 x

10 60

Cross multiply

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3 years ago

Please help very urgent

Alik [6]

Answer:

32, <u>16</u>, <u>8</u>, <u>4</u>, <u>2</u>, 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:

$a_{1} = 32 • (\frac{1}{2})^{1-1}$

$a_{1} = 32 • (\frac{1}{2})^{0}$

$a_{1} = 32 • 1$

$a_{1} = 32$

$a_{6} = 32 • (\frac{1}{2})^{6-1}$

$a_{6} = 32 • (\frac{1}{2})^{5}$

$a_{6} = 32 • \frac{1}{32}$

$a_{6} = 1$

3 0

3 years ago

Read 2 more answers