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xxTIMURxx [149]
2 years ago
15

The function f(t)=-5t^2+15t+50 models the approximate height in feet of an object t seconds after it is launched. how many secon

ds does it take for the object to hit the ground?
Mathematics
1 answer:
Cloud [144]2 years ago
4 0
F(t) can be factored to find the zeros.
.. f(t) = -5(t -5)(t +2) . . . . f(t) = 0 for t=-2, t=5

The object hits the ground after 5 seconds.
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Solve this using square roots and show your work please<br> 4x^2-256=0
mamaluj [8]

Answer:

x = ± 8

Step-by-step explanation:

Given

4x² - 256 = 0 ( add 256 to both sides )

4x² = 256 ( divide both sides by 4 )

x² = 64 ( take the square root of both sides )

x = ± \sqrt{64} = ± 8

Solutions are x = - 8, x = 8

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Find all solutions in the interval [0,2pi)<br> sec^2x + 2tanx = 3
Marysya12 [62]
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3 years ago
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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