Answer:
Step-by-step explanation:
The roots are very clear on the graph. I have left them unlabeled so that you can put the two points in.
The points are (-4,0) and (-2,0)
The graph was done on desmos which you can look up. The box in the upper left corner was filled with
y = x^2 + 6x + 8
Answer:
3
Step-by-step explanation:
This experiment has four groups. Each of these groups have six observations
Total number of observations = 4 x 6 = 24
Treatment degrees of freedom can be gotten by using this formula
Df1 = K - 1
= 4-1
= 3
Therefore for this question, the treatment degrees of freedom is given as 3 and the correct answer option is b.
Triangle JKL has vertices J(−2, 2) , K(−3, −4) , and L(1, −2) .
Rule: (x, y)→(x + 8, y + 1 )
J’ (-2, 2) → (-2 + 8, 2 + 1 ) → (6, 3 )
K’ (-3, -4) → (-3 + 8, -4 + 1 ) → (5, -3 )
L’ (1, -2) → (1 + 8, -2 + 1 ) → (9, -1)
J’ (6,3)
K’ (5,-3)
L’ (9,-1)
Hope this helps!
C is correct
process of elimination will help you prove your answer
a- numbers trailing a decimal before a nonzero digit are not significant
b- zeros between nonzero are significant
d-the zero before the 6 is not significant
<span> Percents as decimals are that number divided by 100. The rule is to move the decimal 2 places to the left.
.9
.3
1.159
.09
.07
.65
.003
4.45
edit:
None need to be rounded so do you mean given to the thousandth place? In that case the thousandth is 3 decimal slots places to the left so like this:
.900
.300
1.159
.090
.070
.650
.003
4.450</span>