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Shtirlitz [24]
3 years ago
9

Ani saves at Rp 800,000.00 on a bank that gives a single interest rate of 16% per year. By the time it was taken,

Mathematics
1 answer:
klasskru [66]3 years ago
7 0
8 is correct so A would be correct. Have a good day! If this is correct like pleaseee

Explanation:
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Make a equation for the scatter plot line
aalyn [17]

Answer:

y = -7.2x + 540

Step-by-step explanation:

to find the slope I used the x-and-y intercepts: (0, 540) and (75, 0)

slope = 540/-75 or -7.2

5 0
3 years ago
What is the area of a triangle that has a height of 6 feet and a base length of 7 feet?
frez [133]

Answer:

area = 21 feet^2

Step-by-step explanation:

Area of a triangle is 1/2(base*height)

1/2(6*7) = area

1/2(42) = area

area = 21 feet^2

6 0
3 years ago
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If one half of a number is 8 less than two thirds of the number what is the number
ludmilkaskok [199]

Answer:

48

Step-by-step explanation:

4 0
3 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
What is the sum of the complex numbers 6+5i and 8+3i2? (NOTE: −1‾‾‾√=i )
Ahat [919]

Answer:

11 + 5i

Step-by-step explanation:

6 + 5i + 8 + 3i² =

= 6 + 8 + 3(-1) + 5i

= 14 - 3 + 5i

= 11 + 5i

4 0
3 years ago
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