PROBLEM 2 What is the greatest common factor of 15 and 30?

Mathematics

2 answers:

seropon [69]2 years ago

7 0

Answer: 15

Step-by-step explanation:

Liono4ka [1.6K]2 years ago

4 0

Answer:

$15 = 3 \times 5 \\ 30 = 3 \times 5 \times 2 \\ \\ \: \: \: \: \: \: \: = 3 \times 5 \\ \: \: \: \: \: \: \: = 15$

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A vehicle travels 57 kilometers north and then travels 26 kilometers south. What is the current position of the vehicle? (2 poin

solniwko [45]

Answer:

C. 31 kilometers north of its starting location

Step-by-step explanation:

<em>Please see attached a rough sketch of the situation for your reference</em>.

Step one:

Displacement North= 57km

Displacement South= 26km

Required

The final displacement

The current position is attained by subtracting 26km from 57km

=57-26= 31km

Therefore the current position is

C. 31 kilometers north of its starting location

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Discuss the continuity of the function on the closed interval.Function Intervalf(x) = 9 − x, x ≤ 09 + 12x, x > 0 [−4, 5]The f

quester [9]

Answer:

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.