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postnew [5]
3 years ago
9

HELP ME PLZ WITH GEOMETRY!!!!

Mathematics
1 answer:
postnew [5]3 years ago
5 0
X=7.1 you create a ratio 7/10=5/x then cross multiply and get 50=7x then divide 50 by 7
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What is the missing step in this proof?
Arada [10]
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>
3 0
3 years ago
Read 2 more answers
What is the probability that 2 randomly selected months have 31 days?
aniked [119]

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%

6 0
3 years ago
Read 2 more answers
What is the product of 4 2/3 and 11 1/4
mihalych1998 [28]

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


7 0
3 years ago
No links nor files. number answer only
katen-ka-za [31]

Answer: The volume is 20 cubic feet

The Width is 1/3 feet

Hope I helped

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Determine whether the sequences converge.
Alik [6]
a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}

Notice that

\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}

So as n\to\infty you have a_n\to0. Clearly a_n must converge.

The second sequence requires a bit more work.

\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}

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

Monotonicity is often easier to establish IMO. You can do so by induction. When n=2, you have

a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1

Assume a_k\ge a_{k-1}, i.e. that a_k=\sqrt{2a_{k-1}}\ge a_{k-1}. Then for n=k+1, you have

a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k

which suggests that for all n, you have a_n\ge a_{n-1}, so the sequence is increasing monotonically.

Next, based on the fact that both a_1=\sqrt2=2^{1/2} and a_2=2^{3/4}, 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

a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}
a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}

and so on. We're getting an inkling that the explicit closed form for the sequence may be a_n=2^{(2^n-1)/2^n}, 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, a_1=2^{1/2}. Let's assume this is the case for n=k, i.e. that a_k. Now for n=k+1, we have

a_{k+1}=\sqrt{2a_k}

and so by induction, it follows that a_n for all n\ge1.

Therefore the second sequence must also converge (to 2).
4 0
3 years ago
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