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romanna [79]
2 years ago
10

AXYZ is reflected across the y-axis. What are the endpoints of segment YZ?

Mathematics
1 answer:
Paul [167]2 years ago
5 0

Answer:

I'm gonna head with B forgive me I'd I'm wrong

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What is 27+3÷ 3 +4-1+72.4-34.24+6-4+75÷5 <br> Answer this fast please!
Dimas [21]

Answer:

86.16

Step-by-step explanation:

27+3÷ 3 +4-1+72.4-34.24+6-4+75÷5

PEMDAS

Multiply and divide first

3/3 is 1   and 75/5 = 15

Replace these into the equation

27+1 +4-1+72.4-34.24+6-4+15

Then add and subtract from left to right

31 +72.4-34.24+6-4+15

69.16+6-4+15

86.16

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It's possible to build a triangle with a side lengths of 5, 5, and 10
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Yeah, base is ten and the two sides are 5...
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A school district's superintendent wants to administer a test of life skills to a sample of students. The superintendent has amp
tangare [24]

Answer:

-Convenience Sampling

Step-by-step explanation:

Convenience Sampling is a sampling method that is employed simply because the samples are easy to reach or access. It is a non-probability system of sampling and not a very strong method of sampling. For a serious research such as the one being conducted by the Superintendent, who has ample time to perform the research, this is not a very good sampling method. A Convenience style of research would mean that he can just walk into a cafeteria with students and administer the test to them.

Since he has detailed information of the students, he should choose any of the other mentioned sampling techniques in the options provided to randomly select the subjects. <em>This would help to</em> prevent bias, and enable him to make generalizations with the result of the research.

6 0
3 years ago
What is the limit of f (×) as x approaches - infinty
tatyana61 [14]
If the degree of numerator and denominator are equal, then limit will be leading coefficient of numerator divided by the leading coefficient of denominator.

So then the limit would be 3/1 = 3.

Alternatively,

\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}=\displaystyle \lim_{x\to\infty}\dfrac{3x^2+6}{x^2-4}\cdot\dfrac{1/x^2}{1/x^2}=\lim_{x\to\infty}\dfrac{3+\frac6{x^2}}{1-\frac4{x^2}} = \dfrac{3+0}{1-0}=\boxed{3}

Hope this helps.
5 0
3 years ago
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