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77julia77 [94]
3 years ago
7

Work out the total surface area of this hemisphere which has a radius of 10 cm.

Mathematics
1 answer:
julsineya [31]3 years ago
6 0

Answer:

200π cm^2

Step-by-step explanation:

Surface area of an emisphere = radius^2 * π * 2 = 10^2 * π * 2 = 200π cm^2

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From the set {7, 21, 63}, use substitution to determine which value of x makes the inequality true.
valina [46]

Answer:

63

Step-by-step explanation:

Assume, instead of > there was a =

In that case, x ÷ 7 = 3, then x would be 21. But we want a value that will be exceed this 21. So the only option you have is 63.

4 0
3 years ago
Nitrogen and oxygen are the most abundant gases in Earth’s atmosphere. A nitrogen molecule is 0.00000000003 meters larger than a
RSB [31]
0.00000000003 \rightarrow 3 \times 10^{-11}



\textrm {Answer:}

\bullet \textrm{The exponent is negative.}

\bullet \textrm{The exponent represents the number of places the decimal moves.}
3 1
3 years ago
Read 2 more answers
(6h+2)+(-7h-8) simplify?
lawyer [7]

Answer:

-h - 6

Step-by-step explanation:

(6h + 2) + (-7h - 8) <== remove parenthesis

6h + 2 - 7h - 8 <== combine like terms

6h - 7h + 2 - 8

-h - 6

Hope this helps!

4 0
2 years ago
Find the cube roots of 27(cos 330° + i sin 330°)
Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

<u>DeMoivre's Theorem</u>

Let z=r(cos\theta+isin\theta) be a complex number in polar form, where n is an integer and n\geq1. If z^n=r^n(cos\theta+isin\theta)^n, then z^n=r^n(cos(n\theta)+isin(n\theta)).

<u>Nth Root of a Complex Number</u>

If n is any positive integer, the nth roots of z=rcis\theta are given by \sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}} where the nth roots are found with the formulas:

  • \sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr] for degrees (the one applicable to this problem)
  • \sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr] for radians

for  k=0,1,2,...\:,n-1

<u>Calculation</u>

<u />z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)

<u>First cube root where k=2</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]

<u>Second cube root where k=1</u>

\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]

<u>Third cube root where k=0</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]

4 0
2 years ago
joe purchased his dream car in 2012 for $30000 just three years later. the book value was only $21000 find a linear equation mod
mrs_skeptik [129]
Y=30,000-21,000= 9,000
4 0
4 years ago
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