Find all solutions tan^2 θ − 1 = 0

Mathematics

2 answers:

Kruka [31]3 years ago

5 0

20-1=19............................................................

nignag [31]3 years ago

4 0

20-1=19 hop this helps :)

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The height of a tennis ball tossed into the air is modeled by h(x) = 40x â€" 16x 2, where x is elapsed time in seconds. During w

egoroff_w [7]

The time interval x >0.5 and x > 2.04 will the tennis ball be at least 15 feet above the ground.

Given that,

The height of a tennis ball tossed into the air is modeled by,

$\rm h(x) = 40x - 16x^2$

Where x is elapsed time in seconds.

We have to determine,

During what time interval will the tennis ball be at least 15 feet above the ground?

According to the question,

The height of a tennis ball tossed into the air is modeled by,

$\rm h(x) = 40x - 16x^2$

Where x is elapsed time in seconds.

Then,

When the tennis ball be at least 15 feet above the ground,

$h(x) = 15$

Substitute the value of h(x) in the equation,

$\rm h(x) = 40x - 16x^2\\\\ 15 = 40x - 16x^2\\\\ 16x^2-40x+15=0$

Factorize the equation for finding the time interval will the tennis ball be at least 15 feet above the ground is,

$\rm 16x^2-40x+15= 0\\\\x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x = \dfrac{-(-40)\pm \sqrt{(-40)^2-4 \times 16 \times 15}}{2 \times 16}\\\\x = \dfrac{-(-40)\pm \sqrt{1600- 960}}{32}\\\\x = \dfrac{40\pm \sqrt{640}}{32}\\\\x = \dfrac{40 + 25.29}{32} \\\\x = \dfrac{40 + 25.29}{32} \ and \ x = \dfrac{40 - 25.29}{32}\\\\x = \dfrac{65.29}{32} \ and \ x = \dfrac{14.71}{32}\\\\x = 2.04 \ and \ x = 0.45$