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VLD [36.1K]
3 years ago
10

What is the value of 5k to the power of 2 - 6 when k equals 9​ A. 84 B. 375 C. 399 D. 2019​

Mathematics
1 answer:
masha68 [24]3 years ago
6 0

Answer:

D

Step-by-step explanation:

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WILL MARK BRAINLIEST ONLY HAVE 10 MINUTES AND DOUBLE POINTS
allochka39001 [22]

Hey there!

They are both correct, and their expressions are equivalent.'

9(S + T) + 45 = 9S + 9T + 45

Have a terrificly amazing day!

8 0
2 years ago
Answering these questions
Rudiy27
1.\\ a)\ y=2x-3\ \ \ |subtract\ y\ from\ both\ sides\\\\\boxed{2x-y-3=0}\\\\b)\ y=\dfrac{1}{3}x-\dfrac{2}{5}\ \ \ |multiply\ both\ sides\ by\ 15\\\\15y=5x-3\cdot2\ \ \ |subtract\ 15y\ from\ both\ sides\\\\\boxed{5x-15y-6=0}


2.\\a)\ 5x-y+3=0\ \ \ |add\ y\ to\ both\ sides\\\\\boxed{y=5x+3}\\\\d)\ 7x+2y-3=0\ \ \ |subtract\ 7x\ from\ both\ sides\\\\2y-3=-7x\ \ \ |add\ 3\ to\ both\ sides\\\\2y=-7x+3\ \ \ \ |divide\ both\ sides\ by\ 2\\\\\boxed{y=-3.5x+1.5}


3.\\y=25x+100\ \ \ \ |subtract\ y\ from\ both\ sides\\\\\boxed{25x-y+100=0}
6 0
3 years ago
What is the sequence of 8,2,0,2,8,18​
antoniya [11.8K]

Answer:

Step-by-step explanation:

32

7 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
What is 0.063 in scientific notation?
Pavel [41]
Remember that you must change this number so that it is between 1 and 10, and then multiply it by a power of 10.

So first, you have to move the decimal point over 2 places to the right.

Since you are moving to the right, the exponent on ten will be negative; since you moved 2 decimal places, you know the exponent will be -2.

The answer is 6.3 x 10^-2.

Hope this helps!
7 0
3 years ago
Read 2 more answers
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