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madam [21]
3 years ago
6

What is Z x-2, when x=2 and z=2

Mathematics
2 answers:
bija089 [108]3 years ago
7 0

Answer:

2

Step-by-step explanation:

2x2-2=2

yawa3891 [41]3 years ago
4 0
2 x 2-2 because everytbing is just two
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Show your work<br> −126 = 14k
Yuliya22 [10]

Answer: k = -9

Step-by-step explanation:

-126 = 14k

-------  -------

 14     14

-126 divided by 14 = -9

k = -9

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PLEASE HELP ME!! WILL GIVE BRAINLIEST PLEASEEE!!
Nikolay [14]
AAS postulate is the answer to this question
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The x-intercept of the line whose equation is 2x - 3y = 6 is
kati45 [8]
2x - 3y = 6
To find the x-intercept, plug zero in for y and solve.
2x - 3(0) = 6
2x = 6
divide both sides by 2
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6 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
A cafeteria used 5 gallon of milk in preparing lunch. How many 1-quart containers of milk did the cafeteria use?
nadya68 [22]
20 quartsg allons
4 quarts a gallon
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