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givi [52]
3 years ago
7

*IN A TEST AND MUST ANSWER MUST BE BEFORE 9:20AM AT 12/7/18

Mathematics
1 answer:
dexar [7]3 years ago
6 0

Answer:

X) 7, 14, 28, 35

Y) 1, 2, 4, 5

Step-by-step explanation:

NO TIME TURN THE TEST IN!!!


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Solve by the square root method 2(5x-7)^2-1=15
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X=11/5 x=2.2 is slicing for x
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3 years ago
Evaluate the limit of sequence:
mr_godi [17]

Rationalize both the numerator and denominator. Given

\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}

we can rationalize it by introducing conjugates of the numerator and denominator:

\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d} \cdot \dfrac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d} \\\\ = \dfrac{\left(\sqrt a\right)^2 - \left(\sqrt b\right)^2}{\left(\sqrt c\right)^2-\left(\sqrt d\right)^2} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b} \\\\ = \dfrac{a-b}{c-d} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b}

Then the limit is equivalent to

\displaystyle \lim_{n\to\infty} \frac{(n+3)-n}{(n+1)-n} \cdot \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = 3 \lim_{n\to\infty} \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n}

For the remaining expression, divide through uniformly by \sqrt n:

\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = \dfrac{\sqrt{1+\frac1n} + 1}{\sqrt{1+\frac3n}+1}

As <em>n</em> goes to infinity, the remaining terms containing <em>n</em> converge to 0, leaving

\dfrac{\sqrt{1}+1}{\sqrt1+1} = \dfrac22 = 1

making the overall limit 3.

8 0
2 years ago
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Answer:

1- 1,092

2- 2,448

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Step-by-step explanation:

I got it correct

3 0
3 years ago
Read 2 more answers
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larisa [96]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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It is 50 degrees.

Since you already know two are 90 and 40 add them together and you get 130.

A triangle has 180 degrees so minus 130 from that

180-130=50
5 0
2 years ago
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