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4 years ago

6 million rounded to the nearest thousand

Mathematics

1 answer:

LuckyWell [14K]4 years ago

7 0

Answer:

Already Rounded (6,000,000)

Step-by-step explanation:

Locate the thousands spot.

6,000,000

2. Check next door.

The hundreds place has a zero digit. Since zero is less than 5, we round down.

Since 6,000,000 rounded to the nearest thousand is the same number, we can say that it's already rounded.

Brainilest Appreciated.

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Write the phrase as a numerical expression; the product of 5 and 7

steposvetlana [31]

Answer: 12 = 7 + 5

Step-by-step explanation:

12 = 7 + 5

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3 years ago

-5x-3.5 < 6.5 -5x <10 x >2

Studentka2010 [4]

Answer:

x=3 ??i think

Step-by-step explanation:

cuz it makes sense like that

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3 years ago

Ginny’s favorite cookie recipe requires 1 1/2 cups of sugar to make 24 cookies. How much sugar does Ginny need to make 36 of the

Delicious77 [7]

Ginny needs $2\frac{1}{4}$ cups of sugar to make 36 cookies.

Step-by-step explanation:

Given,

Sugar cups required for 24 cookies = $1\frac{1}{2}=\frac{3}{2}\ cups$

Ratio of sugar cups to cookies = $\frac{3}{2}:24$

Let,

x be the number of cups needed for 36 cookies.

Ratio of sugar cups to cookies = $x:36\\$

Using proportion;

Ratio of sugar cups to cookies :: Ratio of sugar cups to cookies

$\frac{3}{2}:24::x:36$

Product of mean = Product of extreme

$24*x=\frac{3}{2}*36\\24x=54$

Dividing both sides by 24

$\frac{24x}{24}=\frac{54}{24}\\x=\frac{9}{4}\\x=2\frac{1}{4}$

Ginny needs $2\frac{1}{4}$ cups of sugar to make 36 cookies.

Keywords: fraction, proportion

Learn more about fractions at:

#LearnwithBrainly

7 0

4 years ago

Please solve the problem with steps

Debora [2.8K]

Answer:

Infinite series equals 4/5

Step-by-step explanation:

Notice that the series can be written as a combination of two geometric series, that can be found independently:

$\frac{3^{n-1}-1}{6^{n-1}} =\frac{3^{n-1}}{6^{n-1}} -\frac{1}{6^{n-1}} =(\frac{1}{2})^{n-1} -\frac{1}{6^{n-1}}$

The first one: $(\frac{1}{2})^{n-1}$ is a geometric sequence of first term ( $a_1$ ) "1" and common ratio (r) " $\frac{1}{2}$ ", so since the common ratio is smaller than one, we can find an answer for the infinite addition of its terms, given by: $Infinite\,Sum=\frac{a_1}{1-r} = \frac{1}{1-\frac{1}{2} } =\frac{1}{\frac{1}{2} } =2$

The second one: $\frac{1}{6^{n-1}}$ is a geometric sequence of first term "1", and common ratio (r) " $\frac{1}{6}$ ". Again, since the common ratio is smaller than one, we can find its infinite sum: