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8_murik_8 [283]
3 years ago
12

Express the complex number in trigonometric form.

Mathematics
1 answer:
Maru [420]3 years ago
7 0
-5i can be written as 0 + (-5)i
It is in the form a+bi where a = 0 and b =-5
So the point (a,b) is (0,-5)

The distance from the origin to this point is 5 units, therefore r = 5. This is the magnitude. 

The angle is 270 degrees as shown in the attached image. You start on the positive x axis and rotate until you reach the point (0,-5)

This is why the answer is choice A) 5(cos(270) + i*sin(270))

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Which plant is nonvascular? <br> I didn't see an option for science so I chose math :^
vladimir1956 [14]

Answer:

#4

These are the nonvascular plants or bryophytes (mosses, liverworts, and hornworts), the seedless vascular plants (clubmosses and ferns including, horsetails, club mosses, and whisk ferns), gymnosperms (conifers, cycads, Ginkgo, and gnetophytes), and angiosperms, or flowering plants.

4 0
1 year ago
Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a
8_murik_8 [283]

Answer:

  • a) 3/5·((-2)^n + 4·3^n)
  • b) 3·2^n - 5^n
  • c) 3·2^n + 4^n
  • d) 4 - 3 n
  • e) 2 + 3·(-1)^n
  • f) (-3)^n·(3 - 2n)
  • g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form a[n]=r^n, then it will satisfy ...

  r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}

Rearranging and dividing by r^{n-2}, we get the quadratic ...

  r^2-c_1r-c_2=0

The quadratic formula tells us values of r that satisfy this are ...

  r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

  a[n]=pr_1^n+qr_2^n

We can find p and q by solving the initial condition equations:

\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

These have the solution ...

p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}

_____

Using these formulas on the first recurrence relation, we get ...

a)

c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n

__

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

  a[n]=(p+qn)r^n

The initial condition equations are now ...

\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]

and the solutions for p and q are ...

p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}

__

Using these formulas on problem (d), we get ...

d)

c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n

__

And for problem (f), we get ...

f)

c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0
2 years ago
The weight of meteorite A is 5 times the weight of meteorite B. If the sum of their weights is 132 tons, find the weight of each
sladkih [1.3K]

Answer:

A: 105.6 B: 26.40

Step-by-step explanation:

if you divide 132 by 5, you get Meteorite B, 26.40, then minus that from 132, and you get Meteorite A, 105.6.

To check, do 26.40x5 and you should get 132.

3 0
3 years ago
Select the expression that has a sum of seventy-four hundredths.
Alex_Xolod [135]
It should be D

Explanation:
“Four-tenths (0.4) plus thirty-four hundredths (0.34).”

0.4
0.34
——-
0.74

Hope this helps! :D
4 0
2 years ago
Zachary bought 8.25 pounds of red, white, and blue candies for his Fourth of July party. If 2.5 pounds of the candies are red, a
Dahasolnce [82]

Answer:

2 pounds

Step-by-step explanation:

pounds of white candy: 8.25 - (2.5 + 3.75) = 8.25 - 6.25 = 2

3 0
3 years ago
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