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Serhud [2]
3 years ago
10

Help Help Help Help Brainly please... If you were to plan a mission to Mars, what are the key phases that you need to include in

your mission?
Mathematics
2 answers:
siniylev [52]3 years ago
4 0

Answer:

<em>1.</em><em> Determine the Mission Objective. A mission objective has to be clear, measurable, and believable. </em>

<em>2.</em><em> Identify the Threats. ... </em>

<em>3.</em><em> Identify Your Available Resources. ... </em>

<em>4.</em><em> Evaluate the Lessons Learned. ... </em>

<em>5</em><em>. Determine Courses of Action/Tactics. ... </em>

<em>6.</em><em> Plan for Contingencies.</em>

garik1379 [7]3 years ago
3 0
I would go with what that person said
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Can someone plzzz help me with numbers 15 and 16 olzzz show work cuz I don’t understand plzz
sertanlavr [38]
Pythagoras says
a squared = b squared + c squared so
xsquared= 21 squared +17 squared
x squared = 441 + 34
x squared = 475
x = squared root of 475
x= 21.79mi
6 0
3 years ago
Which of the following equations was used to graph the line shown B).y=x÷2 D).y=x+2​
amid [387]

Answer:

B

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
3 years ago
Helppppppppppppppppppppppppppppppppppppppppp
Kamila [148]

Answer:

30 %

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Will someone help me find the missing angles please
Umnica [9.8K]
1.
the angle should add up to 360
so
360-110-90=160
that is the answer

2.
the angle has to add up to 90
so
90-57=33
7 0
3 years ago
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