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dmitriy555 [2]
4 years ago
7

Cosine, tangent, and sine (topic) how do you solve this (work shown please)

Mathematics
1 answer:
damaskus [11]4 years ago
3 0
Sine θ = \frac{opposite}{hypotenuse}
Cosine θ = \frac{adjacent}{hypotenuse}
Tangent θ = \frac{opposite}{adjacent}

∴ Sin x  =   \frac{e}{f}
   Cos x =  \frac{d}{f}
   Tan x =  \frac{e}{d}
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Solve each inequality for x. (Enter your answers using interval notation.)
olganol [36]

Answer:

(a) ln(x) = 0

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(b) e^x > 2

Then ln2 < x < ∞

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ln(a + b) + ln(a - b) - 5ln(c)

= ln[(a² - b²)/c^5]

Step-by-step explanation:

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(a) ln(x) < 0

=> x < e^(0)

x < 1 ....................................(1)

But the logarithm of 0 is 1, and the logarithm of negative numbers are undefined, we can exclude the values of x ≤ 0.

In fact the values of x that satisfy this inequalities are between 0 and 1.

Therefore, we write:

0 < x < 1

(b) e^x > 2

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and must be finite.

We write as:

ln2 < x < ∞

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3x - 17 = e^5

3x = 17 + e^5

x = (1/3)(17 + e^5)

= 55.1377197

Third Part.

We need to write

ln(a + b) + ln(a - b) - 5ln(c)

as a single logarithm.

ln(a + b) + ln(a - b) - 5ln(c)

= ln(a + b) + ln(a - b) - ln(c^5)

= ln[(a + b)(a - b)/(c^5)]

= ln[(a² - b²)/c^5]

3 0
3 years ago
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