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3 years ago

What is the range of the function y = 2 sin x

Mathematics

2 answers:

mestny [16]3 years ago

5 0

Answer:

[-2, 2] {y | -2 ≤ y ≤ 2}

Step-by-step explanation:

alexandr402 [8]3 years ago

4 0

Answer:

Range: [-2, 2] {y | -2 ≤ y ≤ 2}

Step-by-step explanation:

he given function is y = 2sinx

We have to find the range of the given function.

Since at y-axis sine function has the variance between -1 to 1 so range of y = sinx is [-1, 1]

and if the amplitude of sine function is 2 then sine function will vary from -2 to 2 which implies that range of y = 2 sinx will be [-2, 2]

So range of y = 2sinx is [-2, 2]

Read more on Brainly.com - brainly.com/question/4419238#readmore

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3 years ago

What is the area of the trapezoid shown? trapezoid A = ______ units²?

notka56 [123]

Answer:

74.1 units²

Step-by-step explanation:

Area of the trapezoid shown = ½×(AD + BC)×BE

Use distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to find AD, BC, and BE.

✍️Distance between A(-9, 4) and D(9, 1):

$AD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(9 -(-9))^2 + (1 - 4)^2}$

$= \sqrt{(18)^2 + (-3)^2}$

$= \sqrt{(324 + 9} = \sqrt{333}$

$AD = 18.2 units$ (nearest tenth)

✍️Distance between B(-4, -3) and C(2, -4):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 -(-4))^2 + (-4 -(-3))^2}$

$= \sqrt{(6)^2 + (-1)^2}$

$= \sqrt{(36 + 1} = \sqrt{37}$

$BC = 6.1 units$ (nearest tenth)

✍️Distance between B(-4, -3) and E(-3, 3):

$BE = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-3 -(-4))^2 + (3 -(-3))^2}$

$= \sqrt{(1)^2 + (6)^2}$

$= \sqrt{(1 + 36} = \sqrt{37}$

$BE = 6.1 units$ (nearest tenth)

✔️Area of the trapezoid:

Area = ½×(AD + BC)×BE

Plug in the values into the formula

Area = ½×(18.2 + 6.1)×6.1

Area = ½×(24.3)×6.1

Area = 74.115 ≈ 74.1 units² (nearest tenth)

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3 years ago