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Tanya [424]
3 years ago
15

What is the domain of the following list of ordered pairs? (-3, -2), (1, 5), (-2, 4)

Mathematics
1 answer:
Marina86 [1]3 years ago
6 0
X∈{-3,-2,1}, as there are only points, only certain set numbers can be included within the domain
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Is a negative and a positive equal a negative or positive in mutiplication
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If you multiply a negative number by a positive number, you get a negative product. if you were to multiply two positive or two negative numbers, you'd get positive products.
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ASAP! I will give you branilest
Yuliya22 [10]

Answer:

1.6cm

Step-by-step explanation:

80/50=1.6 cm

Good Luck!

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What is the solution to this inequality?
Ksivusya [100]
The answer is D. 

X/-5 = 2
x =2 * -5
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3 years ago
Tabitha has a deck of cards numbered 1−10. She picks one card, puts it back in the deck and then chooses a second card. What is
Valentin [98]
First compute the whole number of the possibilities :
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Please help me with these
Alex Ar [27]
When we approach limits, we are finding values that are infinitesimally approaching this x-value. Essentially, we consider the approximate location that this root or limit appears. This is essential when it comes to taking Calculus, and finding the limit or rate of change of a function.

When we are attempting limits questions, there are several tests we attempt first.

1. Evaluate the limit by substituting the value of the x-value as it approaches the value (direct evaluation of a limit)
2. Rearrangement of the function, such that we can evaluate the limit.
3. (TRIGONOMETRIC PROPERTIES)
\lim_{x \to 0} (\frac{sinx}{x}) = 1
\lim_{x \to 0} (\frac{tanx}{x}) = 1
4. Using L'Hopital's Rule for indeterminate limits, such as 0/0, -infinity/infinity, or infinity/infinity.

For example:

1) \lim_{x \to 0}\frac{\sqrt{x} - 5}{x - 25}

We can do this using the first and second method.
<em>Method 1: Direct evaluation:</em>

Substitute x = 0 to the function.
\frac{\sqrt{0} - 5}{0 - 25}
= \frac{-5}{-25}
= \frac{1}{5}

<em>Method 2: Rearranging the function
</em>

We can see that x - 25 can be rewritten as: (√x - 5)(√x + 5)
By rewriting it in this form, the top will cancel with the bottom easily, and our limit comes out the same.

\lim_{x \to 0}\frac{(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}
= \lim_{x \to 0}\frac{1}{(\sqrt{x} + 5)}}
= \frac{1}{5}

Every example works exactly the same way, and by remembering these criteria, every limit question should come out pretty naturally.
8 0
3 years ago
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