Given that Relationship B has a lesser rate than Relationship A and that the graph representing Relationship A is a f<span><span>irst-quadrant graph showing a ray from the origin through the points
(2, 3) and (4, 6) where the horizontal axis label is Time in weeks and the vertical axis
label is Plant growth in inches.</span>
The rate of relationship A is given by the slope of the graph as follows:

To obtain which table could represent Relationship B, we check the slopes of the tables and see which has a lesser slope.
For table A.
Time (weeks) 3 6 8 10
Plant growth (in.) 2.25 4.5 6 7.5

For table B.
Time (weeks) 3 6 8 10
Plant growth (in.) 4.8 9.6 12.8 16
</span><span><span>

</span>
For tabe C.
Time (weeks) 3 4 6 9
Plant growth (in.) 5.4 7.2 10.8 16.2
</span><span>
For table D.
Time (weeks) 3 4 6 9
Plant growth (in.) 6.3 8.4 12.6 18.9</span>
<span>

</span>
Therefore, the table that could represent Relationship B is table A.
Answer:
$39
Step-by-step explanation:
Using the given formula :
I = P × R × T
I = 30 × 10/100 × 3
I = $9
She have = 30 + 9 = $39
Answer:
Step-by-step explanation:
the answer is 14
.2 is 2/10 while .5 is 1/2. 1/2 is much larger than 1/5 (which is 2/10 reduced).
Answer:
=
$
518.01
Explanation:
compound interest formula $A = P*(1+R)^n#
P
=
$
400
,
r
=
.09
,
n
=
3
Substituting the values, we get A =
400
⋅
(
1.09
)
3
=
$
518.01
From my alt account from, https://socratic.org/questions/how-much-would-400-invested-at-9-interest-compounded-continuously-be-worth-after