A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, the distance to the cat appears to be A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, the distance to the cat appears to be Equal to the actual distance. Less than the actual distance. More than the actual distance.

Hence A fish in an aquarium see the cat more than actual distance.

Step-by-step explanation:

Given:

A fish in aquarium with flat silde looks out to cat .

To find :

Appearance of fish to the cat.

Solution:

Now this problem is related to the refractive index of 2 mediums

So cat is  in air medium and a fish in water i.e. aquarium

R.I of water =1.33

R.I of air =1.00

We know the incident ray ,reflected ray,refracted ray and normal .

When a incident ray enter in denser medium it bends towards  normal.

But it diverges outward direction and goes beyond the actual object.

(Refer the Attachment).

Hence A fish in an aquarium see the cat more than actual distance.

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: