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QveST [7]
3 years ago
5

How do I do this!! I’m so confused

Mathematics
1 answer:
nydimaria [60]3 years ago
4 0
So each example is still in the ratio of the triple on the left.
**note: the largest factor will alway be the hypotenuse

a.1. 15, 20, ? in a 3, 4, 5 ratio
15/5 = 3
20/5 = 4
x/5 = 5
x = 5×5 = 25

b.1. ?, 24, 26 in a 5, 12, 13 ratio
26/2 = 13
24/2 = 12
x/2 = 5
x = 2×5 = 10

c.1. 70, 240, ? in a 7, 24, 25 ratio
70/10 = 7
240/10 = 24
x/10 = 25
x = 10×25 = 250

d.1. 24, 45, ? in an 8, 15, 17 ratio
24/3 = 8
45/3 = 15
x/3 = 17
x = 3×17 = 51
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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x &gt; 0 [−4, 5]The f
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It is continuous since \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)

Step-by-step explanation:

We are given that the function is defined as follows f(x) = 9-x, x\leq 0 and f(x) = 9+12x, x>0 and we want to check the continuity in the interval [-4,5]. Note that this a piecewise function whose only critical point (that might be a candidate of a discontinuity)  x=0 since at this point is where the function "changes" of definition. Note that 9-x and 9+12x are polynomials that are continous over all \mathbb{R}. So F is continous in the intervals [-4,0) and (0,5]. To check if f(x) is continuous at 0, we must check that

\lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x) (this is the definition of continuity at x=0)

Note that if x=0, then f(x) = 9-x. So, f(0)=9. On the same time, note that

\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} 9-x = 9. This result is because the function 9-x is continous at x=0, so the left-hand limit is equal to the value of the function at 0.

Note that when x>0, we have that f(x) = 9+12x. In this case, we have that

\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 9+12x = 9. As before, this result is because the function 9+12x is continous at x=0, so the right-hand limit is equal to the value of the function at 0.

Thus, \lim_{x\to 0^{-}} = f(0) = \lim_{x \to 0^{+} f(x)=9, so by definition, f is continuous at x=0, hence continuous over the interval [-4,5].

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4 years ago
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