5+10-20 asdfghjklzxcvbnmmmmmmmmmmmqwertyuiop
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 with Step-by-step explanation:
We are given that:
tanA = 0.45
tanA=
=
⇒ =0.45
⇒ =
⇒ =
⇒ AC= 20
Hence, by pythagoras theorem
AB²=AC²+BC²
AB²=20²+9²
AB²=481
⇒ AB=21.9 units
Hence, the approximate length of AB is:
22 units