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Nitella [24]
3 years ago
11

A delivery fee for tuber eats was 10

Mathematics
1 answer:
yaroslaw [1]3 years ago
5 0
$10.63
the work is shown below
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Factor the common factor out of 8x3 + 12x. please help asap.
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So 8x3 = 24

24 + 12x
the common factor is 12
so you divide everything by 12
24 ÷ 12 = 2
12x ÷ 12 = x

final answer: 12 (2+x)
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Match the part of the circle with the term that best describes it.​
nordsb [41]
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Bridget rolls a fair dice 84 times.<br> How many times would Bridget expect to roll an odd number?
cluponka [151]

Answer:

42 times

Step-by-step explanation:

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What is the solution to the trigonometric inequality 2sin(x)+3&gt;sin ^2(x) over the interval
navik [9.2K]

The intervals that satisfy the given trigonometric Inequality are; 0 ≤ x < 3π/2 and 3π/2 < x ≤ 2π

<h3>How to solve trigonometric inequality?</h3>

We are given the trigonometric Inequality;

2 sin(x) + 3 > sin²(x)

Rearranging gives us;

sin²(x) - 2 sin(x) - 3 < 0

Factorizing this gives us;

(sin(x) - 3)(sin(x) + 1) < 0

Thus;

sin(x) - 3 = 0 or sin(x) + 1 = 0

sin(x) = 3 or sin(x) = -1

sin(x) = 3 is not possible because sin(x) ≤ 1.

Thus, we will work with;

sin(x) = -1 for the interval 0 ≤ x ≤ 2π radians.

Then, x = sin⁻¹(-1)

x = 3π/2.

Now, if we split up the solution domain into two intervals, we have;

from 0 ≤ x < 3π/2, at x = 0. Then;

sin²(0) - 2 sin(0) - 3

= 0² - 0 - 3

= -3 < 0

Thus, the interval 0 ≤ x < 3π/2 is true.

From 3π/2 < x ≤ 2π, take x = 2π. Then;

sin²(2π) - 2 sin(2π) - 3

= 0² - 0 - 3

= -3 < 0

Thus, the interval 3π/2 < x ≤ 2π is also true.

Read more about trigonometric inequality at; brainly.com/question/27862380

#SPJ1

3 0
1 year ago
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