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Alinara [238K]
3 years ago
12

Find the ratio in simplest form. 30:6

Mathematics
2 answers:
Norma-Jean [14]3 years ago
7 0
5:1 divide both by 6

krok68 [10]3 years ago
4 0
Should be 5:1, I'm not sure though.
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Find the fourth roots of 16(cos 200° + i sin 200°).
NeTakaya

Answer:

<em>See below.</em>

Step-by-step explanation:

To find roots of an equation, we use this formula:

z^{\frac{1}{n}}=r^{\frac{1}{n}}(cos(\frac{\theta}{n}+\frac{2k\pi}{n} )+\mathfrak{i}(sin(\frac{\theta}{n}+\frac{2k\pi}{n})), where k = 0, 1, 2, 3... (n = root; equal to n - 1; dependent on the amount of roots needed - 0 is included).

In this case, n = 4.

Therefore, we adjust the polar equation we are given and modify it to be solved for the roots.

Part 2: Solving for root #1

To solve for root #1, make k = 0 and substitute all values into the equation. On the second step, convert the measure in degrees to the measure in radians by multiplying the degrees measurement by \frac{\pi}{180} and simplify.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(0)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(0)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))

z^{\frac{1}{4}} = 2(sin(\frac{5\pi}{18}+\frac{\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{4}))

<u>Root #1:</u>

\large\boxed{z^\frac{1}{4}=2(cos(\frac{19\pi}{36}))+\mathfrack{i}(sin(\frac{19\pi}{38}))}

Part 3: Solving for root #2

To solve for root #2, follow the same simplifying steps above but change <em>k</em>  to k = 1.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(1)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(1)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{2\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{2\pi}{4}))\\

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{\pi}{2}))\\

<u>Root #2:</u>

\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{7\pi}{9}))+\mathfrak{i}(sin(\frac{7\pi}{9}))}

Part 4: Solving for root #3

To solve for root #3, follow the same simplifying steps above but change <em>k</em> to k = 2.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(2)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(2)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{4\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{4\pi}{4}))\\

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\pi))+\mathfrak{i}(sin(\frac{5\pi}{18}+\pi))\\

<u>Root #3</u>:

\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{23\pi}{18}))+\mathfrak{i}(sin(\frac{23\pi}{18}))}

Part 4: Solving for root #4

To solve for root #4, follow the same simplifying steps above but change <em>k</em> to k = 3.

z^{\frac{1}{4}}=16^{\frac{1}{4}}(cos(\frac{200}{4}+\frac{2(3)\pi}{4}))+\mathfrak{i}(sin(\frac{200}{4}+\frac{2(3)\pi}{4}))

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{6\pi}{4}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{6\pi}{4}))\\

z^{\frac{1}{4}}=2(cos(\frac{5\pi}{18}+\frac{3\pi}{2}))+\mathfrak{i}(sin(\frac{5\pi}{18}+\frac{3\pi}{2}))\\

<u>Root #4</u>:

\large\boxed{z^{\frac{1}{4}}=2(cos(\frac{16\pi}{9}))+\mathfrak{i}(sin(\frac{16\pi}{19}))}

The fourth roots of <em>16(cos 200° + i(sin 200°) </em>are listed above.

3 0
3 years ago
The line segment joining A(6, 3) to B(–1, –4) is doubled in length by having half its length added to each end. Find the coordin
mr Goodwill [35]

Answer:

  • (9.5, 6.5) and (-4.5, -7.5)

Step-by-step explanation:

Let the extended points be A' and B' and add the point M as midpoint of AB

<u>Coordinates of M are:</u>

  • ((6 - 1)/2, (3-4)/2) = (2.5, -0.5)

Now point A is midpoint of A'M and point B is midpoint of MB'

<u>Finding the coordinates using midpoint formula:</u>

  • A' = ((2*6 - 2.5),(2*3 - (-0.5)) = (9.5, 6.5)
  • B' = ((2*(-1) - 2.5), (2*(-4) - (-0.5)) = (-4.5, -7.5)

4 0
3 years ago
Suppose y = 48 + 3(2n - 1) is an explicit representation of an arithmetic sequence for integer values n ≥ 1. Find the xth partia
omeli [17]
a_n=48+3(2n-1)

The formula of the sum of the arithmetic sequence:
S_n=\dfrac{a_1+a__n}{2}\cdot n
calculate:
a_1=48+3(2\cdot1-1)=48+3=51
substitute
S_n=\dfrac{51+48+3(2n-1)}{2}\cdot n=\dfrac{99+6n-3}{2}\cdot n=\dfrac{96+6n}{2}\cdot n=3n^2+48n
Your answer is:
\boxed{f(x)=3x^2+48x}

5 0
3 years ago
Read 2 more answers
PLS answer the question below. I will mark brainliest!!
sesenic [268]

Answer:

5 triangles maybe?

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
the unit digitsof a two numberis 1 less than twice the ten digit. if the sum of tje digits is 9, find the number.
Amiraneli [1.4K]
Are you sure this is written correctly? Any number whose digits add to 9 is divisible by 9.  For a 2 digit number then, your choices are
18, 27, 36, 45, 54, 63, 72, and 81.  None of these has the property that for the 2 digit number AB that B=2A-1.  Solving your question algebraically is possible but you get fractional digits.
5 0
3 years ago
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