Catherine and Lance open a savings account at the same time. Catherine deposits $100 initially and adds $20 per week. Lance depo
2 answers:
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The borders are shown in the picture attached.
As you can see, starting with border 1, we have 6 daises (white squares) surrounded by 10 tulips (colored squares). Through Jerry's expression we expected:
<span>8(b − 1) + 10 =
</span>8(1 − 1) + 10 =
0 + 10 =
10 tulips.
When considering border 2, we expect:
<span>8(b − 1) + 10 =
</span>8(2 − 1) + 10 =
8 + 10 =
<span>18 tulips.
Indeed, we have the 10 tulips from border 1 and 8 additional tulips, for a total of 18 tulips.
Then, consider border 3, we expect:
</span><span>8(b − 1) + 10 =
</span>8(3 − 1) + 10 =
16 + 10 =
26<span> tulips.
Again, this is correct: we have the 10 tulips used in border 1 plus other 16 tulips, for a total of 26.
Therefore, Jerry's expression is correct.</span>
As you can see, starting with border 1, we have 6 daises (white squares) surrounded by 10 tulips (colored squares). Through Jerry's expression we expected:
<span>8(b − 1) + 10 =
</span>8(1 − 1) + 10 =
0 + 10 =
10 tulips.
When considering border 2, we expect:
<span>8(b − 1) + 10 =
</span>8(2 − 1) + 10 =
8 + 10 =
<span>18 tulips.
Indeed, we have the 10 tulips from border 1 and 8 additional tulips, for a total of 18 tulips.
Then, consider border 3, we expect:
</span><span>8(b − 1) + 10 =
</span>8(3 − 1) + 10 =
16 + 10 =
26<span> tulips.
Again, this is correct: we have the 10 tulips used in border 1 plus other 16 tulips, for a total of 26.
Therefore, Jerry's expression is correct.</span>
Answer:
Option B. t=5/3w ;1/3
Step-by-step explanation:
We are told that varies Inversely with
thus generally speaking
∝
since they are inverted, <u>otherwise</u> if they were proportionally varied then
∝
. This means that there is a constant value (
) for which
is inversely proportional to
and can be mathematically expressed as:
Eqn. (1)
Now since we are given the values of and
, we can plug them in Eqn. (1) and find our constant of proportionality
as follow:
Now that we have our constant we can find the new value for the second value of
as follow:
Therefore based on the options give, we can see that Option B. is correct since
and for
, then