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2 years ago

For problems 3 and 4, find the missing side of the triangle. Leave answers in simplest radical form.

Mathematics

1 answer:

maria [59]2 years ago

7 0

Answer: 3. 4 $\sqrt{13}$ 4. $\sqrt{225}$

Step-by-step explanation:

8^2 + 12^3 = c^2

64 + 144 = c^2

208 = c^2

√208

$4\sqrt{13}$

The final answer is the square root of 208

8^2 + b^2 = 17^2

64 + b^2 = 289

-64 64

b^2 = 225

$\sqrt{225}$

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Answer:

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3 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

Eginning at 15,000 feet, a commercial jetliner begins climbing at a rate (in feet per minute) given by f(t)=e0.4t+8 f ( t ) = e

nydimaria [60]

Answer:

16,126 ft

Step-by-step explanation:

Using the net change theorem and letting s(t) represent the aircraft's position, and s(0) be the aircraft's position at time t = 0, s(0) = 15000 ft and s(15 be the aircraft's position at time, t = 15 minutes respectively, then,

s(15) - s(0) = ∫₀¹⁵f(t)dt

s(15) - s(0) = ∫₀¹⁵[ $e^{0.4t} + 8$ ]dt

s(15) - 15000 = [ $\frac{e^{0.4t}}{0.4} + 8t$ ]₀¹⁵

s(15) - 15,000 ft = $\frac{e^{0.4X15}}{0.4} + 8X15 - \frac{e^{0.4 X 0}}{0.4} + 8(0)\\\frac{e^{6}}{0.4} + 120 - \frac{e^{0}}{0.4} + 0\\\frac{403.43}{0.4} + 120 - \frac{1}{0.4} + 0\\1008.57 + 120 - 2.5\\1126.07$

s(15) = 15,000 ft + 1126.07 ft

s(15) = 16,126.07 ft

s(15) ≅ 16,126 ft to the nearest foot

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2 years ago