1. AD/DB=CE/EB
given
2. AD/DB+1=CE/EB+1
Addition Property of Equality
3. AD+DB/DB=CE+EB/EB
using common denominators
4. AB = AD + DB
CB = CE + EB
segment addition
5. AB/DB=CB/EB
Substitution Property of Equality
6.ABC DBE
Reflexive Property of Congruence
7. ABC ~ DBE
SAS similarity criterion
<span>8. </span>
Corresponding angles of similar triangles are congruent.
9. DE|| AC
<span>If the corresponding angles formed by </span>
Answer:
31.8%
Step-by-step explanation:There are 7 months that have 31 days so,
The odds the first selected month has 31 days is 7/12
The next month there are only 6 months left with 31 days of only 11 months to choose from
so the odds for the second are 6/11
To calculate the odds they BOTH have 31 months you multiply the two odds:
(7/12) x (6/11) = 42/132 = .318 so 31.8%
First you have to get rid of the mixed numbers, so multiply the whole numbers with the denominators and add the numerators so that way you get:
14/3 times 45/4....then cross simplify if possible and finally you can multiply the numerators and the denominators...
7/1 times 15/2 = 105/2....simplify it and get: 52 1/2
Answer: The volume is 20 cubic feet
The Width is 1/3 feet
Hope I helped
Step-by-step explanation:

Notice that

So as

you have

. Clearly

must converge.
The second sequence requires a bit more work.

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then

will converge.
Monotonicity is often easier to establish IMO. You can do so by induction. When

, you have

Assume

, i.e. that

. Then for

, you have

which suggests that for all

, you have

, so the sequence is increasing monotonically.
Next, based on the fact that both

and

, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.
We have


and so on. We're getting an inkling that the explicit closed form for the sequence may be

, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.
Clearly,

. Let's assume this is the case for

, i.e. that

. Now for

, we have

and so by induction, it follows that

for all

.
Therefore the second sequence must also converge (to 2).