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NemiM [27]
3 years ago
6

I will give brainliest please help me asap! Simplify the expression. Make sure you have all positive exponents.​

Mathematics
1 answer:
Lemur [1.5K]3 years ago
7 0

Step-by-step explanation:

\frac{ {x}^{3} }{ {x}^{8} }  =  {x}^{3 - 8}  \\  =  {x}^{ - 5}  \\  =  \frac{1}{ {x}^{5} }

To leave answer in positive exponents, just leave the answer inverse of the negative exponents.

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Step-by-step explanation:

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The question is somewhat poorly posed because the equation doesn't involve <em>θ</em> at all. I assume the author meant to use <em>x</em>.

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By definition of secant and cosecant,

1/cos(<em>x</em>) = 1/sin(<em>x</em>)

Multiply both sides by sin(<em>x</em>) :

sin(<em>x</em>)/cos(<em>x</em>) = sin(<em>x</em>)/sin(<em>x</em>)

As long as sin(<em>x</em>) ≠ 0, this reduces to

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By definition of tangent,

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Solve for <em>x</em> :

<em>x</em> = arctan(1) + <em>nπ</em>

<em>x</em> = <em>π</em>/4 + <em>nπ</em>

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5 0
3 years ago
Q.5(b) The population {(P) in millions} of a country is estimated by the function, P=125e0.035t, t = time measured in years sinc
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Answer:

Hello,

Step-by-step explanation:

Q.5(b) The population {(P) in millions} of a country is estimated by the function, P=125e0.035t, t = time measured in years since 1990. (a) what is the population expected to equal in year 2000 (b) determine the expression for the instantaneous rate of change in the population (c) what is the instantaneous rate of change in the population expected to equal in year 2000.

P(t)=125*e^{0.035*t}\\a)\\P(2000)=125*e^{0.035*(2000-1990)}=177.38...\\\\b)\\P'(t)=125*0.035*e^{0.035*t}\\\\c)\\\\P'(2000)=125*0.035*e^{0.035*(2000-1990)}=6.2084...

5 0
3 years ago
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Answer:

Step-by-step explanation:

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Answer:

563.76

here ya go

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