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Georgia [21]
2 years ago
15

I need help in this!

Mathematics
1 answer:
In-s [12.5K]2 years ago
3 0

For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that a is such that

\displaystyle\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)\iff\lim_{x\to0}ax=\lim_{x\to0}x^2-3x

Obviously, both limits are 0, so g is indeed continuous at x=0.

Now, for g to be differentiable everywhere, its derivative g' must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals:

g'(x)=\begin{cases}a&\text{for }x0\end{cases}

and at this time, we don't know what's going on at x=0, so we omit that case. We want g' to be continuous, so we require that

\displaystyle\lim_{x\to0^-}g'(x)=\lim_{x\to0^+}g'(x)\iff\lim_{x\to0}a=\lim_{x\to0}2x-3

from which it follows that a=-3.

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rusak2 [61]

Answer:

P_{m}=(6,10.5,9)

Step-by-step explanation:

The mid point can be found with the formula

P_{m}=(\frac{x_{1}+x_{2} }{2},\frac{y_{1} +y_{2} }{2} ,\frac{z_{1}+z_{2}  }{2} )

The given coordinates are P(5,10,8) and Q(7,11,10).

Replacing coordinates in the formula, we have

P_{m}=(\frac{5+7}{2},\frac{10+11 }{2} ,\frac{8+10}{2} )=(\frac{12}{2},\frac{21 }{2} ,\frac{18}{2} )\\P_{m}=(6,10.5,9)

Therefore, the mid point of the segment PQ is P_{m}=(6,10.5,9)

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3 years ago
The translation that moves a figure down 4, to the right 2 is
lorasvet [3.4K]
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P and q are two arithmetic sequences. The first four terms of sequence P are 2,6,10,14. The nth term of sequence Q is 145-3n. Th
Tomtit [17]

Answer:

r = 21

Step-by-step explanation:

nth of sequence P :

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x=70 vertical opposite

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Read 2 more answers
(20 POINTS!!!)
frozen [14]

Answer:

1170 ft³

Step-by-step explanation:

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= 1170

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