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3 years ago

I really dont get it halp ;-;

Mathematics

2 answers:

erik [133]3 years ago

6 0

Answer:

Step-by-step explanation:

This would be easier to explain to you on a graph, but I shall do the best I can in text format.

You know what a complex plane looks like, right? Just like a number line but with a vertical number line added on at the point (0). This makes four quadrants, you know the drill.

So each quadrant holds all values from on of the four positive/negative combinations, (1, 1i), (1, -1i), (-1, -1i), and (-1, 1i), in that order.

So this means that all the pairs with both positive values are in quadrant I, all the pairs with a positive rational number but a negative irrational number are in II, etc.

If you have a graph, you can simply count along the number line to find the answer, so in this case, starting as always at (0, 0), move 6 units to the right and then eight units down, where you end up in the lower righthand quadrant, which is II.

Oof I know that was worded badly but I hope it's not totally useless to you <3

You're doing great with the test, Gem, I'm sure you'll do just fine <:

Advocard [28]3 years ago

4 0

Answer:

HOPE IT HELP IT WILL YOU❣️

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3 years ago

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Find the critical points, domain endpoints, and local extreme values for the function

aliina [53]

Answer:

a) $x = -3$ , b) $x = 0$ , $x = -6$ , c) $x = 0$ , d) $x = -6$

Step-by-step explanation:

a) Let derive the function:

$f'(x) = \frac{10\cdot x \cdot (x+3)-5\cdot x^{2}}{25\cdot (x+3)^{2}}$

$f'(x)$ is undefined when denominator equates to zero. The critical point is:

$x = -3$

b) $f'(x) = 0$ when numerator equates to zero. That is:

$10\cdot x \cdot (x+3) - 5\cdot x^{2} = 0$

$10\cdot x^{2}+30\cdot x -5\cdot x^{2} = 0$

$5\cdot x^{2} + 30\cdot x = 0$

$5\cdot x \cdot (x+6) = 0$

This equation shows two critical points:

$x = 0$ , $x = -6$

c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that $x = 0$ corresponds to a local maximum. (See Attachment)

d) By plotting the function, it is evident that $x = -6$ corresponds to a local minimum. (See Attachment)