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timofeeve [1]
3 years ago
11

A coordinate grid with 2 lines. The first line labeled f(x) passes through (negative 3, 3), (0, 2), and (3, 1). The second line

labeled g(x) passes through the points at (negative 3, 0) and (0, 2)
What is the solution to the system of linear equations?

(–3, 0)
(–3, 3)
(0, 2)
(3, 1)
Mathematics
1 answer:
vesna_86 [32]3 years ago
8 0

Answer:

(0, 2)

Step-by-step explanation:

the answer is (0, 2)

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Answer:

Step-by-step explanation:

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3 years ago
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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 
\lim_{x \rightarrow 4-} f(x) = \lim_{x \rightarrow 4-} (x^2 - c^2) = 16 - c^2

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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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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Answer:

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Step-by-step explanation:

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time taken = 13 minutes and 20 seconds

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Answer:

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