4 years ago

Subtract (3 + 2i) from (-9 - 8i)

Mathematics

1 answer:

zheka24 [161]4 years ago

4 0

Answer:

$-12-10i$ is the required answer.

Step-by-step explanation:

$\left(-9-8i\right)-\left(3+2i\right)\\\\\mathrm{Group\:the\:real\:part\:and\:the\:imaginary\:part\:of\:the\:complex\:number}\\\\=\left(-9-3\right)+\left(-8-2\right)i\\\\=-12-10i$

Best Regards!

You might be interested in

Find two unit vectors orthogonal to a=⟨2,−2,−3⟩a=⟨2,−2,−3⟩ and b=⟨3,2,2⟩b=⟨3,2,2⟩ enter your answer so that the first non-zero c

Ronch [10]

We need the cross product of the two vectors.
a x b
=<2,-2,-3> x <3,2,2>
=
i j k
2 -2 -3
3 2 2
=<-4+6, -(4+9), 4+6>
=<2,-13,10>
The second vector is obtained by reversing the direction, namely <-2,13,-10>
Thus the two vectors are <2,-13,10> and <-2,13,-10>.

8 0

3 years ago

What #3 and 4 plz help me ^w^

serious [3.7K]

#3
6 x 4 - 4
First multiply 6 x 4
24 - 4
———
5
Subtract the top

20
—
5

Then divide it

7 0

4 years ago

What is the domain of the sine and cosine functions?

Sholpan [36]

Answer:

all real numbers

only the range value is with a minimum of -1 and a maximum of 1 unless there's a vertical shift.