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2 years ago

Round this number to the nearest 10,000. 2,874,992

Mathematics

2 answers:

muminat2 years ago

7 0

For this case we have the following number:

2,874,992

We want to round this number to the nearest ten thousand.

We then have the following rule:

1) If the number before the rounding place is less than 5, then the next number does not change.

2) If the number before the rounding place is greater than or equal to 5, then the next number increases by one.

Rounding the number we have:

2,874,992 = 2,870,000

Answer:

The rounded number for the nearest ten thousand is:

2,870,000

Alinara [238K]2 years ago

5 0

The number rounded to the nearest ten thousands 2,870,000.

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A petting zoo owner is building an enclosure for his farm animals. He needs 353 feet of fencing. The supply store sells fencing

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Ares of a rectangular floor of your house is Ax+5x-2. if the length is x+2 units, find A

Snezhnost [94]

Answer:

A=4.

Step-by-step explanation:

Note: In the expression of area the first term should be $Ax^2$ .

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$Area=length\times width$

Since length of the rectangle is $(x+2)$ , therefore $(x+2)$ is a factor of $Ax^2+5x-2$ and the value of $Ax^2+5x-2$ is 0 at x=-2.

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2 years ago

A skier is trying to decide whether or not to buy a season ski pass. A daily pass costs $68. A season ski pass costs $300. The

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2 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

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2 years ago