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tekilochka [14]
3 years ago
13

Cual es la forma más simple al efectuar la operación indicada en la expresión (-3) (-12) ?

Mathematics
1 answer:
gulaghasi [49]3 years ago
7 0

Answer:

Cual es la forma más simple al efectuar la operación indicada en la expresión (-3) (-12) ?  this means What is the simplest way to perform the operation indicated in the expression (-3) (-12)?

Step-by-step explanation:

Multiply -3 by -12 and get 36. hope this was helpful.

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Answer:

3^12

Step-by-step explanation:

Simplify the expression, 3^10/3^-2.

Next, solve for the sum. which will be 3^12.

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42 + 3y ≤ 54

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Which statement is correctly written as a conditional statement?
bekas [8.4K]

The statement that correctly written as a conditional statement is; A number, such as 10, is a composite number because it is even. so the correct option is D.

<h3>What are natural numbers, rational numbers, integers and irrational numbers?</h3>

Natural numbers are: 1, 2, 3, ....

Integer numbers are: ...., -2, -1, 0, 1, 2, ... (so it includes positive and negative natural number, and 0 )

Rational numbers are numbers which can be written in the form of a/b  

where a and b are integers.

Irrational numbers are those real numbers which are not rational numbers.

Here, Know that all natural numbers are integers, all integers are rational numbers. That means, natural numbers are not irrational.

A composite number is a number divisible by another number than 1 and the number itself.

To find example, a sum of two composite numbers, which is not a composite number, is needed.

The statement that correctly written as a conditional statement is; A number, such as 10, is a composite number because it is even.

Hence so the correct option is D.

Learn more about composite number;

brainly.com/question/19312484

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Estimates of the number of cells in the human body vary over an order of magnitude.​ Indeed, the precise number varies from one
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a) 2.00 × 10¹⁰;  b) 2.00 × 10¹³

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\text{No. of cells} =\text{1 cm}^{3} \times\left( \dfrac{\text{1 m}}{\text{100 cm}}\right )^{3} \times \left (\dfrac{1 \, \mu \text{m} }{10^{-6}\,\text{ m}}\right )^{3} \times  \dfrac{\text{1 cell}}{\text{50 $\,\mu$m}^{3}} = \mathbf{2.00 \times 10^{10}} \textbf{ cells}

b) No. of cells in 1 L

\text{No. of cells} =  \text{1 L} \times \dfrac{\text{1000 cm}^{3}}{\text{1 L}} \times \dfrac{2 \times 10^{10} \text{ cells}}{\text{1 cm}^{3}} = \mathbf{2.00 \times 10^{13}} \textbf{ cells}

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