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iogann1982 [59]
3 years ago
13

The circle x2+y2=36 is translated 5 Units left and 4 units up where is the center of the new circle after the translation

Mathematics
1 answer:
Andrej [43]3 years ago
6 0
The general equation of a circle is given by:
(x-a)^2+(x-b)^2=r^2
where:
(a,b) is the center
r is the radius

given the equation:
x^2+y^2=36
it means that the equation is centered at (0,0) with radius of 6 units. Thus a translation of 5 units to the left and 4 units up, will change the new center to
(-5,4)
thus the equation will be:
(x+5)^2+(y-4)^2=36

Answer: (x+5)^2+(y-4)^2=36
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Komok [63]

Answer:

a) The critical points are x = 3 and x = -6.

b) f is decreasing in the interval (-\infty, -6)

f is increasing in the intervals (-6,3) and (3,\infty).

c) Local minima: x = -6

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Step-by-step explanation:

(a) what are the critical points of f?

The critical points of f are those in which f^{\prime}(x) = 0. So

f^{\prime}(x) = 0

(x-3)^{2}(x+6) = 0

So, the critical points are x = 3 and x = -6.

(b) on what intervals is f increasing or decreasing? (if there is no interval put no interval)

For any interval, if f^{\prime} is positive, f is increasing in the interval. If it is negative, f is decreasing in the interval.

Our critical points are x = 3 and x = -6. So we have those following intervals:

(-\infty, -6), (-6,3), (3, \infty)

We select a point x in each interval, and calculate f^{\prime}(x).

So

-------------------------

(-\infty, -6)

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---------------------------

(-6,3)

f^{\prime}(2) = (2-3)^{2}(2+6) = (1)(8) = 8

f is increasing in the interval (-6,3).

------------------------------

(3, \infty)

f^{\prime}(4) = (4-3)^{2}(4+6) = (1)(10) = 10

f is increasing in the interval (3,\infty).

(c) At what points, if any, does f assume local maximum and minima values. ( if there is no local maxima put mo local maxima) if there is no local minima put no local minima

At a critical point x, if the function goes from decreasing to increasing, it is a local minima. And if the function goes from increasing to decreasing, it is a local maxima.

So, for each critical point is this problem:

At x = -6, f goes from decreasing to increasing.

So x = -6, f assume a local minima value

At x = 3, f goes from increasing to increasing. So, there it is not a local maxima nor a local minima. So, there is no local maxima for this function.

4 0
3 years ago
What is the answer?
Alex73 [517]

Answer:

see explanation

Step-by-step explanation:

k=3, giving 3, 4, 5 as the three integers.

He should have used k + (k+2) + (k+4) = 4k

which gives k = 4 and the correct answer is

4, 6, 8

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