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3 years ago

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

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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