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3 years ago

What is the value of 5^3i^9?

Mathematics

1 answer:

Yakvenalex [24]3 years ago

4 0

Answer:

$\displaystyle 125^i$

Step-by-step explanation:

Extended Information on the Complex Number System

$\displaystyle \sqrt{-1} = i \\ -1 = i^2 \\ -i = i^3 \\ 1 = i^4\:[all\:multiples\:of\:four]$

I am joyous to assist you at any time.

All work is shown above. ↑

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What's worth more a 100 dollars or 350 quarters?

Ronch [10]

Maybe 200 dollars, 500 dollars and so on

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3 years ago

What is the value of v? 4/5(v−7)=2

Nostrana [21]

4/5(v - 7) = 2...distribute the 4/5 thru the parenthesis
4/5v - 28/5 = 2 ...multiply everything by 5
4v - 28 = 10...add 28 to both sides
4v = 10 + 28
4v = 38
v = 38/4 reduces to 19/2 or 9 1/2 <==

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3 years ago

Read 2 more answers

88) 4 times a number decreased by 8 equals 40 A) x=80 B) x = 40 C) x=8 D) x = 12

dimulka [17.4K]

D=12
Because if you write down the equation 4x-8=40 and plug in 12. 4 times 12 is 48
. 48-8 equals 40

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3 years ago

17-(5-p)=2(5p-16) Step by step explanation please

12345 [234]

Answer:

p= 44 /9

(44 over 9)

Step-by-step explanation:

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3 years ago

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

3 years ago

Read 2 more answers