SUPER URGENT: What is the range of y = sinx in the interval – π ≤ x ≤ 0? -1 ≤ y ≤ 1 -1 ≤ y ≤ 0 0 ≤ y ≤ 1 y ≤ 1

Question

SUPER URGENT: What is the range of y = sinx in the interval – π ≤ x ≤ 0?
-1 ≤ y ≤ 1
-1 ≤ y ≤ 0
0 ≤ y ≤ 1
y ≤ 1

in progress 0
Thiên Hương 1 month 2021-08-30T21:06:46+00:00 1 Answers 0 views 0

Answers ( )

    0
    2021-08-30T21:08:26+00:00

    Answer:

    -1 ≤ y ≤ 0

    Step-by-step explanation:

    here, we simply want to know the range of values for sin x over the given interval

    The range of values for sin x are simply the values in which y will take over the given interval of x

    let’s start with the value sine (-pi)

    We have this as;

    0

    And the value of sin (0) = 0

    Recall, between -pi and 0, we have (-pi/2)

    so sin (-pi/2) = -1

    Thus, we have the correct range of values as;

    -1 ≤ y ≤ 0

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )