What is the range of this data set?

Mathematics

2 answers:

Svet_ta [14]3 years ago

4 0

Answer:the answer is 15.8

Step-by-step explanation:

Vinil7 [7]3 years ago

4 0

Answer:

15.8 i had this on a math test:/

Step-by-step explanation:

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Find the complex fourth roots of 81(cos(3π/8)+isin(3π/8)). a) Find the fourth root of 81. b) Divide the angle in the problem by

WARRIOR [948]

Answer:

The answer is below

Step-by-step explanation:

Let a complex z = r(cos θ + isinθ), the nth root of the complex number is given as:

$z_1=r^{\frac{1}{n} }(cos(\frac{\theta +2k\pi}{n} )+isin(\frac{\theta +2k\pi}{n} )),\\k=0,1,2,.\ .\ .,n-1$

Given the complex number z = 81(cos(3π/8)+isin(3π/8)), the fourth root (i.e n = 4) is given as follows:

$z_{k=0}=81^{\frac{1}{4} }(cos(\frac{\frac{3\pi}{8} +2(0)\pi}{4} )+isin(\frac{\frac{3\pi}{8} +2(0)\pi}{4} ))=3[cos(\frac{3\pi}{32} )+isin(\frac{3\pi}{32})] \\z_{k=0}=3[cos(\frac{3\pi}{32} )+isin(\frac{3\pi}{32})]\\\\z_{k=1}=81^{\frac{1}{4} }(cos(\frac{\frac{3\pi}{8} +2(1)\pi}{4} )+isin(\frac{\frac{3\pi}{8} +2(1)\pi}{4} ))=3[cos(\frac{19\pi}{32} )+isin(\frac{19\pi}{32})] \\z_{k=1}=3[cos(\frac{19\pi}{32} )+isin(\frac{19\pi}{32})]\\\\$

$z_{k=2}=81^{\frac{1}{4} }(cos(\frac{\frac{3\pi}{8} +2(2)\pi}{4} )+isin(\frac{\frac{3\pi}{8} +2(2)\pi}{4} ))=3[cos(\frac{35\pi}{32} )+isin(\frac{35\pi}{32})] \\z_{k=2}=3[cos(\frac{35\pi}{32} )+isin(\frac{35\pi}{32})]\\\\z_{k=3}=81^{\frac{1}{4} }(cos(\frac{\frac{3\pi}{8} +2(3)\pi}{4} )+isin(\frac{\frac{3\pi}{8} +2(3)\pi}{4} ))=3[cos(\frac{51\pi}{32} )+isin(\frac{51\pi}{32})] \\z_{k=3}=3[cos(\frac{51\pi}{32} )+isin(\frac{51\pi}{32})]$

3 0

3 years ago

Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte

Mashcka [7]

Answer:

f'(x) > 0 on $(-\infty ,\frac{1}{e})$ and f'(x)<0 on $(\frac{1}{e},\infty)$

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

$f'(x)>0$

To find its decreasing interval :

$f'(x)$

2) Then let's find the critical point of this function:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0$

2.2 Solving for x this equation, this will lead us to one critical point since x' is not defined for Real set, and x'' $=\frac{1}{e}$ ≈0.37 for e≈2.72