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2 years ago

10

What is the square of the imaginary number 3i?

2 answers:

Gwar [14]2 years ago

8 0

Answer:

-9

Step-by-step explanation:

Illusion [34]2 years ago

5 0

−
9
Explanation:
(
3
i
)
2
=
9
i
2
i
2
=
−
1
,
therefore,
9
i
2
=
9
⋅
(
−
1
)
=
−
9

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Triangle EFG in which segment EF measures 3 units and segment FG measures 5 units

Softa [21]

Answer:

Yes

Step-by-step explanation:

It can be 6 units. The rule relating to the sides of the triangle is that you should be able to add two sides and get a length that is longer than the third side. You usually add the lower two numbers because it basically says that it will work for all sides but, if you're still in doubt you just have to compare the numbers.

For Example

3+5 is greater than 6

5+6 is greater than 3

6+3 is greater than 5.

All of these makes sense so 6 can be a possible answer.

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Which group of ordered pairs are on the line given by the equation 5−2=6

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Aleksandr [31]

Answer: Your answer should be 21 I believe.

Step-by-step explanation:

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Elijah buys a jersey that costs $56.00. His bank account balance is now -$6.50. Which math sentence describes how much he owes?

soldier1979 [14.2K]

It would be (B) > $6.00 since he owes $6.50, which is a larger value than $6.00.

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You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

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2 years ago