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abruzzese [7]
3 years ago
13

Equivalent factors of 1/2=2/4= =

Mathematics
2 answers:
gulaghasi [49]3 years ago
5 0
25/50, 50/100, 500,000/1,000,000 are also a few
stiv31 [10]3 years ago
4 0
3/6 and 4/8... is equivalent also to 1/2... If that's what you are looking for...
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Plug in the value of x
Inessa [10]

Answer: yes

Step-by-step explanation:

8 0
3 years ago
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Evaluate the expression 4+−4√2+−1√ and write the result in the form a+bi.
castortr0y [4]

In the expression, the real number a equals 12 and the real number b equals -16.

<h3>How to explain the information?</h3>

It should be noted that the expression given is:

= (4 - 2i)²

Therefore, we need to expand the expression. This will be:

= (4 - 2i)(4 - 2i)

= 16 -8i -8i + 4i²

= 16 -16i +4i²

Substitute -1 for i²

= 16 - 16i + 4(-1)

= 16 - 16i - 4

= 12 - 16i

Therefore, the value of the expression is 12 - 16i.

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brainly.com/question/723406

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Evaluate the expression ( 4 − 2 i )² and write the result in the form a + b i.

8 0
2 years ago
How many 3 digit license plates can be made if the plate starts with a number, ends with a letter and the middle is either one?
svlad2 [7]
Is this a question on your test or something? I know trying out every number and letter would be difficult. There is a way to solve these types of problems. Ask your math teacher, or someone that knows this or has taught this.
5 0
3 years ago
What is 15/22 as a repeating decimal?
andrew11 [14]

Answer:

0.6818

Step-by-step explanation:

6 0
3 years ago
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If lim x-&gt; infinity ((x^2)/(x+1)-ax-b)=0 find the value of a and b
MAXImum [283]

We have

\dfrac{x^2}{x+1}=\dfrac{(x+1)^2-2(x+1)+1}{x+1}=(x+1)-2+\dfrac1{x+1}=x-1+\dfrac1{x+1}

So

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-ax-b\right)=\lim_{x\to\infty}\left(x-1+\frac1{x+1}-ax-b\right)=0

The rational term vanishes as <em>x</em> gets arbitrarily large, so we can ignore that term, leaving us with

\displaystyle\lim_{x\to\infty}\left((1-a)x-(1+b)\right)=0

and this happens if <em>a</em> = 1 and <em>b</em> = -1.

To confirm, we have

\displaystyle\lim_{x\to\infty}\left(\frac{x^2}{x+1}-x+1\right)=\lim_{x\to\infty}\frac{x^2-(x-1)(x+1)}{x+1}=\lim_{x\to\infty}\frac1{x+1}=0

as required.

3 0
3 years ago
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