A: (x + 5i)^2
= (x + 5i)(x + 5i)
= (x)(x) + (x)(5i) + (5i)(x) + (5i)(5i)
= x^2 + 5ix + 5ix + 25i^2
= 25i^2 + 10ix + x^2
B: (x - 5i)^2
= (x + - 5i)(x + - 5i)
= (x)(x) + (x)(- 5i) + (- 5i)(x) + (- 5i)(- 5i)
= x^2 - 5ix - 5ix + 25i^2
= 25i^2 - 10ix + x^2
C: (x - 5i)(x + 5i)
= (x + - 5i)(x + 5i)
= (x)(x) + (x)(5i) + (- 5i)(x) + (- 5i)(5i)
= x^2 + 5ix - 5ix - 25i^2
= 25i^2 + x^2
D: (x + 10i)(x - 15i)
= (x + 10i)(x + - 15i)
= (x)(x) + (x)(- 15i) + (10i)(x) + (10i)(- 15i)
= x^2 - 15ix + 10ix - 150i^2
= - 150i^2 + 5ix + x^2
Hope that helps!!!
Answer:
D. There were no significant effects.
Step-by-step explanation:
The table below shows the representation of the significance level using the two-way between subjects ANOVA.
Source of Variation SS df MS F P-value
Factor A 10 1 10 0.21 0.660
Factor B 50 2 25 0.52 0.6235
A × B 40 2 20 0.42 0.6783
Error 240 5 48 - -
Total 340 10 - - -
From the table above , the SS(B) is determined as follows:
SS(B) = SS(Total)-SS(Error-(A×B)-A)
= 340-(240-40-10)
= 50
A researcher computes the following 2 x 3 between-subjects ANOVA;
k=2
n=3
N(total) = no of participants observed in each group =11
df for Factor A= (k-1)
=(2-1)
=1
df for Factor B = (n-1)
=(3-1)
=2
df for A × B
= 2 × 1
= 2
df factor for total
=(N-1)
=11-1
=10
MS = SS/df
Thus, from the table, the P-Value for all data is greater than 0.05, therefore we fail to reject H₀.
to go from point to point
we go up six units and to the right 1 unit
slope = 6/1
m = 6
1. 5/y=2/3x +3
2. y= 7/1-4
3. y=2/5x8
4. y=-3/4 +1 I could be wrong but I think this is right
I had problems with the same thing when i was in middle school gotta take it step by step
