Answer:
1. In the multiplication of the imaginary parts, the student forgot to square of i. OR
2. The student has only multiplied the real parts and the imaginary parts.
Correct value 
.
Step-by-step explanation:
The given expression is
A student multiplies (4+5i) (3-2i) incorrectly and obtains 12-10i.
Student's mistake can be either 1 or second:
1. In the multiplication of the imaginary parts, the student forgot to square of i.
2. The student has only multiplied the real parts and the imaginary parts.
Which is not correct. The correct steps are shown below.
Using distributive property, we get
![[\because i^2=-1]](https://tex.z-dn.net/?f=%5B%5Cbecause%20i%5E2%3D-1%5D)
Therefore, the correct value of is .
9514 1404 393
Answer:
d = 3√10 ≈ 9.487
Step-by-step explanation:
The distance formula is appropriate.
d = √((x2 -x1)^2 +(y2 -y1)^2)
d = √((-3 -0)^2 +(2 -(-7))^2) = √(9 +81) = √90
d = 3√10
So first create and define your variables:
Z = amount of zebra fish
N = amount of neon tetras
Now create your equations:
2z+2.15n=31.20
z+n=15
This is your system. There are multiple methods to use but in this problem I’ll use the substitution method by simplifying the bottom equation.
2z+2.15n=31.20
z=15-n
Now I’ll plug the bottom equation into the top one.
2(15-n)+2.15n=31.20
And just solve from here.
30-2n+2.15n=31.20
0.15n=1.20
n=8
So he bout 8 neon tetras, and 15-8= 7, so he bought 7 zebra fish
