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ella [17]
3 years ago
8

Find the value of b.A. 14B. 15C. 64D. 289

Mathematics
1 answer:
Dafna11 [192]3 years ago
8 0

9514 1404 393

Answer:

  B.  15

Step-by-step explanation:

From the second equation, ...

  b² = 17² -8²

  b = √(289 -64) = √225

  b = 15

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3 years ago
For what value of c is the function defined below continuous on (-\infty,\infty)?
kozerog [31]
f(x)= \left \{ {{x^2-c^2,x \ \textless \  4} \atop {cx+20},x \geq 4} \right


It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if 
</span><span>\lim_{x \rightarrow 4} \  f(x) = f(4)

</span><span>In notation we write respectively
</span>\lim_{x \rightarrow 4-} f(x) \ \ \ \text{ and } \ \ \ \lim_{x \rightarrow 4+} f(x)

Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence 
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Thus these two limits, the one from above and below are equal if and only if
 4c + 20 = 16 - c²<span> 
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 4c + 20 = 16 - c</span>²

c^2+4c+4=0&#10;\\(c+2)^2=0&#10;\\c=-2

That is to say, if c = -2, f(x) is continuous at x = 4. 

Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers (-\infty, +\infty)

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rodikova [14]

Answer:

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Step-by-step explanation:

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Question 10 is not visible

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