Decompose each velocity in its i and j - coordinates.
1) River's velocity
3.5 m/s, west fo east = 3.5i + 0j
2) Luke's velocity
5m/s, 60° east of north
5* sin(60°) i + 5*cos(60°) j = 4.33 i + 2.5 j
3) Resultant velocity
[3.5i + 0j] + [4.33i + 2.5j] = 7.83i + 2.5j
resultant speed = √[ (7.83)^2 + (2.5)^2 ] = 8.22 m/s
angle = arctan(2.5 / 7.83) = 17.7° north of east (equal to 72.3° east of north).
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 .
Answer:
I think this is the correct answer
Answer: the length of Maria's piece of string is 45 inches.
the length of Katy's piece of string is 39 inches
Step-by-step explanation:
Let x represent the length of Maria's piece of string.
Let y represent the length of Katy's piece of string.
When they put the two pieces of string together end to end, the total length is 84inches. This means that
x + y = 84 - - - - - - - - - - - -1
Maria's string is 6 inches longer than Katy's. This means that
x = y + 6
Substituting x = y + 6 into equation 1, it becomes
y + 6 + y = 84
2y + 6 = 84
2y = 84 - 6 = 78
y = 78/2 = 39
x = y + 6 = 39 + 6
x = 45
One way to understand division is to look at it as repeated
subtraction. When you "divide by" a divisor number, you're
asking "how many times can I subtract this divisor from the
dividend, before the dividend is all used up ?".
Well, if the divisor is ' 1 ', then you're taking ' 1 ' away from the
dividend each time, and the number of times will be exactly
the same as the dividend.
If the divisor is more than ' 1 ', then you subtract more than ' 1 '
from the dividend each time, and the number of times you can
do that is less than the dividend itself.
If the divisor is less than ' 1 ', then you only take away a piece of
' 1 ' each time. You can do that more times than the number in
the dividend, because you only take away a piece each time.
