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VashaNatasha [74]
3 years ago
13

Which of the following is equivalent to the complex number i^8?

Mathematics
2 answers:
sergeinik [125]3 years ago
7 0

Answer:

The correct answer is B.  i⁸ = 1

Step-by-step explanation:

You can work this out by noting that 2 is a factor of 8, and 8 is a multiple of 2.  With that we can say:

i^8\\= (i^2)^4\\= -1^4\\= 1

Maru [420]3 years ago
4 0

Answer:

I think the answer is -1

Step-by-step explanation:

the sequence goes i, -i, 1, -1

Right? >_>

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How do you write the equation 4x + 2y = 16 in function form? Please answer with an explanation!
kramer
F(x)=-2x+8 This is because f(x) is the same as writing y so therefore you have to change the equation to 4x+2y=16 so you have Y=-2x+8
3 0
3 years ago
22 = 5c + 3c - c +8<br>Please solve for c.​
bogdanovich [222]

Answer:2=c

Step-by-step explanation:

22=5c+3c-c+8

22-8=5c+3c-c+8-8

14=5c+3c-c+0

14=5c+3c-c

14=8c-2

14=7c

14/7=7c/7

2=c

Check answer by pluging 2 in for c.

5*2+3*2-2+8

=10+6-2+8

=16-2+8

=14+8

=22

The final answer is 2=c

4 0
3 years ago
Find the missing factor of D that makes the equality true <br><br> -15y^4=(D)(3y^2)<br> D=
sergiy2304 [10]

Answer:

D=-5y^2

Step-by-step explanation:

-15y^4=D*(3y^2)

D=(-15y^4)/(3y^2)=-5y^2

8 0
2 years ago
What is (10^9)^3 written as a single exponent
fenix001 [56]

9514 1404 393

Answer:

  10^27

Step-by-step explanation:

Your calculator can help with this.

  (10^9)^3 = 10^(9·3) = 10^27

__

The applicable rule of exponents is ...

  (a^b)^c = a^(bc)

5 0
3 years ago
Use the substitution u = tan(x) to evaluate the following. int_0^(pi/6) (text(tan) ^2 x text( sec) ^4 x) text( ) dx
Rudiy27
If we use the substitution u = \tan x, then du = \sec^2 {x}\ dx. If you try substituting just u and du into the integrand, though, you'll notice that there's a \sec^2x left over that we have to deal with.

To get rid of this problem, use the identity \tan^2 x + 1 = \sec^2 x and substitute in the left side of the identity for the extra \sec^2x, as shown:

\int\limits^{\pi/6}_0 {tan^2 x \ sec^4 x} \, dx
\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx

From there, we can substitute in u and du, and then evaluate:

\int\limits^{\pi/6}_0 {tan^2 x \ (tan^2 x + 1) \ sec^2 x} \, dx
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^2(u^2 + 1)} \, du
\int\limits^{\frac{1}{\sqrt{3}}}_0 {u^4 + u^2} \, du
= \left.\frac{u^5}{5} + \frac{u^3}{3}\right|_0^\frac{1}{\sqrt{3}}
= (\frac{(\frac{1}{\sqrt{3}})^5}{5} + \frac{(\frac{1}{\sqrt{3}})^3}{3}) - (\frac{(0)^5}{5} + \frac{(0)^3}{3})
= \frac{1}{45\sqrt{3}} + \frac{1}{9\sqrt{3}} = \frac{6}{45\sqrt{3}} = \bf \frac{2}{15\sqrt{3}}


8 0
3 years ago
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