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Snowcat [4.5K]
3 years ago
13

Lim (h->0) (sqrt(25+h)-5)/h.

Mathematics
1 answer:
motikmotik3 years ago
5 0
\lim_{h \to 0}  \frac{ \sqrt{25+h} -5}{h}= \frac{0}{0}
We have to multiply numerator and denominator by: √(25-h)+ 5:
\lim_{h \to 0}  \frac{ \sqrt{25+h}-5}{h}* \frac{ \sqrt{25+h} +5}{ \sqrt{25+h}+5}  = \\  \lim_{h \to 0}  \frac{25+h-25}{h( \sqrt{25+h}+5) }  = \\  \lim_{h \to 0}  \frac{1}{ \sqrt{25+h} +5} =   \\ = \frac{1}{5+5}== 1/10 = 0.1
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The probability that a point is chosen at random in the square is in the blue region is 0.8.

<h3>What is Probability?</h3>

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As we know that the area of the shaded region is the sum of the area of triangle A and the area of triangle B. Therefore, the area of the blue shaded region is,

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brainly.com/question/795909

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4 years ago
Guys need help its surd one <br><img src="https://tex.z-dn.net/?f=5%20%5Csqrt%7B3%20%5Cdiv%202%20%5Csqrt%7B3%20-%20%20%5Csqrt%7B
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Answer:

Step-by-step explanation:

\displaystyle \Large \boldsymbol{} 5\cdot  \sqrt{3:2\sqrt{3-\sqrt{2} } }  =\\\\\\ 5\cdot  \sqrt{\frac{3}{2\sqrt{3-\sqrt{2} } } \cdot \frac{2\sqrt{3+\sqrt{2} } }{2\sqrt{3+\sqrt{2} } } } = \\\\\\5\cdot  \sqrt{\frac{6\sqrt{3+\sqrt{2} } }{4\cdot \sqrt{9-2} } }  =5\sqrt{\frac{6\sqrt{3+\sqrt{2} } }{2\sqrt{7} } }

But if the condition would be like this

\displaystyle \Large \boldsymbol{} 5\cdot  \sqrt{3:2\sqrt{3-\boldsymbol2\sqrt{2} } }  =\\\\\\5\cdot \sqrt{\frac{3}{2\cdot \sqrt{\underbrace{2+1-2\sqrt{2} \cdot \sqrt{1} }_{(\sqrt{2} -1)^2} } }} =\\\\\\5\cdot \sqrt{\frac{3}{2\cdot (\sqrt{2}-1) } } } = \\\\\\ 5\cdot  \sqrt{\frac{3}{2(\sqrt{2}-1 )}  \cdot \frac{\sqrt{2} +1}{\sqrt{2}+1 }}   =5 \sqrt{\frac{3}{2(2-1)} }  =\boxed{5\sqrt{1,5} }

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3 years ago
Please help! Will give Brainliest and 20 POINTS!
jarptica [38.1K]
Your answer would be the first on because when you set up your equation you can find 4 by dividing 42,980 by 10 4 times wich will give you 10^4 when you divide exponents you subtract so iwill look like this....

5x10^6
----------= 1.25x10^2 
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3 0
4 years ago
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