What is the exact value of arccos(− √2/2) ? enter your answer, in terms of π , in the box.

or

Step-by-step explanation:

We want to find the exact value of

.

Let .

This implies that,

Let .

Since the cosine ratio is negative, it means the angle is in the second or third quadrant.

We first of all find

In the second quadrant,

In the third quadrant,

Any of them is correct on the interval,

arccos(-√2/2) = 3π/4 radians

Step-by-step explanation:

As we know Arccos(x)=y means we have to find the cos value of an "Angle y" for which the value is "x".

So as per question arccos(-√2/2) = arccos(-√2/(√2)²) OR (-1/√2)

Now we know that negative value of cos(x) lies in second quadrant.

If Cos x = 1/√2 then x = π/4

But the given value is (-1/√2) so cosx lies in second quadrant and the angle is = π-π/4 = 3π/4

So the arccos(-√2/2) = 3π/4.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

step-by-step explanation: