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d1i1m1o1n [39]
3 years ago
15

What is the distance between the points (-7,3) and (4,-5) on a coordinate plane

Mathematics
2 answers:
Marina86 [1]3 years ago
8 0
B b b b b b b b b b b b b b b
LenKa [72]3 years ago
5 0

Answer:

b because c 13 is too far away for one coordinate were it is not too long and to short

Step-by-step explanation:

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Read 2 more answers
What is the value of x? *
EleoNora [17]

In a circle, the parts of the diagonals multiply to equal each other, so 6·4=2x.

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6 0
2 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 
\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> 
 Or in other words, the limit as x --> 4 of f(x) exists if and only if
 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)

4 0
3 years ago
Help asap !………………………
My name is Ann [436]

Answer:

c

Step-by-step explanation:

dvdywuwhfuwi eurue wuid

4 0
2 years ago
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