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Elena-2011 [213]
3 years ago
12

Neeeed helpppp nowwww!!!!!

Mathematics
1 answer:
siniylev [52]3 years ago
6 0

I think the answer is the second one?

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Is domain of f(x,y) = 1 + (4 -y^2)^1/2 open, closed or neither<br><br> is it bounded or unbounded?
omeli [17]

Answer:

Closed and Bounded.

Step-by-step explanation:

Hi there!

1) Let's start by finding the values in which the function is defined. Remember that this can be rewritten:

f(x,y) = 1 + (4 -y^2)^{\frac{1}{2}} \:id\:est\:f(x,y)=1+\sqrt{(4 -y^2)}

Since every quadratic root are defined for values \geq 0  then, this help us to understand that we need calculate what interval this Domain is:

4-y^{2} \geq 0\\4-y^{2}-4\geq -4\\-y^{2}\geq-4 \\y^{2}\leq4\\-2\leq y\leq 2

D=[-2,2]

2) Graphically speaking, the domain is closed. For the values -2 and 2 are included, and bounded.

Bounded functions have all of their points contained by some circle origin centered. Check it out below.

6 0
4 years ago
Pls who can solve this asap​
andrew-mc [135]

Answer:

the radius    1 < r < 3, so  r = 1.39

Step-by-step explanation:

notice that you have 3 cm radius for the big circle

r radius for small

and you have this rectangle with dimension  6 cm by 8cm

so if you can imagine the radius segments moving into vertical and horizontal position

there are two boxes...  6cm by 6cm  and 2r by 2r,  but  6cm + 2r is not 8cm

diagonal of rectangle = root(8*8 + 6*6)  = 10 cm

diagonals intersect at  ((0 + 8)/ 2,  (0 + 6) / 2)= (4, 3) but center of big circle at (3, 3)

It appears that the point (4, 3) is on the diagonal but does not go through the centers of the circles.  but I do see

slope of diagonal is 3/4  so   if  rise is  2r, then   the run.

(2r/x) = 3/4

(4/3)* 2r = x

equation of diagonal :   y = (3/4)x + 0    if the bottom left is (0,0)

equation of circle r :     (x - (8-r))^2 + (y - (6 - r))^2 = r*r

---------

Use a trig function to find the length of this line segment from the center of the big circle to the intersection of the small circle.  

The triangle  has lengths  1 cm , 3cm and a segment called x

x = line segment from the center of the big circle to the intersection of the small circle.

the angle between 1cm and x  is 180 -  arctan (3/4)  = 180 - 36.8698 deg.

=143.1302 degrees

trig formula:

c^2 =  a^2 + b^2  - 2ab*cos C

c^2 = 1^2 + 3^2 - 2*1*3*cos (81.8698 )

c^2 = 10 - 6 cos (81.8698)

c^2 =  9.15146173

c = 3.0251383  units

ok so we know the diagonal is  10

the distance from the intersection to the corner is  10 - 5 - 3.0251383

= 1.9748616 units

we can say  1.9748  =  root(2) * r

so  that  r =  1.9748 / root(2)

3 0
3 years ago
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