All categories

3 years ago

Which of the following is a complex number?

Mathematics

1 answer:

daser333 [38]3 years ago

5 0

Answer:

C. $3 \sqrt{ \frac{7}{5} } + \sqrt{ - \frac{9}{5} }$

Step-by-step explanation:

Recall that:

$\sqrt{ - 1} = i$

For this particular question, you just have to look for the radical containing a negative sign under the square root i.e a negative radicand.

That will be option C.

This is given as

$3 \sqrt{ \frac{7}{5} } + \sqrt{ - \frac{9}{5} }$

This can be rewritten to reveal the complex part as:

$3 \sqrt{ \frac{7}{5} } + \sqrt{ \frac{9}{5} }i$

Therefore the correct answer us option C

You might be interested in

Please help!!! Ill give brainliest

ale4655 [162]

Answer:

the answer is C

you will calculate the area of the wall and also the window. Then you subtract the area of the window from the wall. the answer is 152 ft square

8 0

2 years ago

WILL GIVE BRAINLIEST IF UR RIGHT!! PLEASE HELP!!

Darina [25.2K]

D. 24
you can simplify it down to y^3/x^2 then you just plug in x and y.

4 0

3 years ago

A sphere and a cylinder have the same radius and height. The volume of the cylinder is 21 meters cubed. A sphere with height h a

Elodia [21]

We have been given that a sphere and a cylinder have the same radius and height. The volume of the cylinder is 21 meters cubed. We are asked to find the volume of the sphere.

We know that height of sphere is equal to its radius. This means that $r=h$ .

$\text{Volume of sphere}=\frac{4}{3}\pi r^3$

$\text{Volume of cylinder}=\pi r^2 h$

Using $r=h$ , we will get:

$\text{Volume of cylinder}=\pi r^2\cdot r$

$\text{Volume of cylinder}=\pi r^3$

Upon comparing volume of sphere and volume of cylinder, we can see that volume of sphere is $\frac{4}{3}$ times volume of cylinder.

$\pi r^3=21$

$\frac{4}{3}\pi r^3=\frac{4}{3}\cdot 21$

$\frac{4}{3}\pi r^3=4\cdot 7$

$\frac{4}{3}\pi r^3=28$

Therefore, the volume of the sphere would be 28 cubic meters.

4 0

3 years ago

Read 2 more answers

◆ Quadratic Equations ◆<br>Please help !

Mila [183]

I'm sure there's an easier way of solving it than the way I did, but I'm not sure what it could be. Never dealt with a problem like this before.

Anyway, I just plugged in and tested. Chose random values for a, b, c, and d, which follow the rule 0 < a < b < c < d:

a = 1
b = 2
c = 3
d = 4

$\sf ax^2+(1-a(b+c))x+abc-d)$

$\sf 1x^2+(1-1(2+3))x+(1)(2)(3)-(4))$

Simplify into standard form:

$\sf x^2+(1-1(5))x+6-4$

$\sf x^2+(1-5)x+2$

$\sf x^2-4x+2$

Use the quadratic formula to solve:

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

For functions in the form of $\sf ax^2+bx+c$ . So in this case:

a = 1
b = -4
c = 2

Plug them in:

$\sf x=\dfrac{4\pm\sqrt{(-4)^2-4(1)(2)}}{2(1)}$

Solve for 'x':

$\sf x=\dfrac{4\pm\sqrt{16-8}}{2}$

$\sf x=\dfrac{4\pm\sqrt{8}}{2}$

$\sf x\approx\dfrac{4\pm 2.83}{2}$

$\sf x\approx 0.59,3.41$

So the answer would be A.

3 0

3 years ago

Determine the number of outcomes in the event. Decide whether the event is a simple event or not. You randomly select one card f

Marat540 [252]

Answer:

No we can consider A as single event. See the explanation below

Step-by-step explanation:

Let's define the event A as:

A=" Select a 4 from a standard deck of 52 cards"

We know that in a standard deck we have 4 different types of 4, spade, heart, diamond and club.

And by definition of simple event we need to have just one possible outcome in the experiment, and on this case we have 4 possible options for event A, so for this reason the event A can't be considered as simpl event.

3 0

3 years ago