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VladimirAG [237]
2 years ago
11

How do you even do this?! i’m so confused

Mathematics
1 answer:
SIZIF [17.4K]2 years ago
7 0

Answer:

D.

Step-by-step explanation:

Pretend the y-axis is a wall, on the left side it is decreasing and on the right it is increasing, so A. and B. aren't correct. C. and D. are left, x<0 means on the left side of the y-axis because x (any number) is less than 0 and on the left side it is decreasing so C. isn't correct so D. is your answer because x>0 means x is greater than 0 so it's on the left side of the y-axis.

I hope this helps!

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Simplify the expression |-11|?
Andrej [43]

Answer:

11

Step-by-step explanation:

| | takes the absolute value of a number so it wouldn't be negative

5 0
2 years ago
What is the inequality?<br> A. x &lt; 2<br> B. x ≤ 2<br> C. y ≥ 2<br> D. x &gt; 2
Lera25 [3.4K]

the answer of this inequality is

A. x<2

3 0
3 years ago
If an experimenter sets equal to .01, then she is defining a "statistically rare" event as an event occurring more than one time
Fynjy0 [20]

Answer:

The correct answer is an event occurring one or fewer times in 100 times if the null hypothesis is true.

Step-by-step explanation:

For a statistically rare event, its probability is relatively small and the event is very unlikely to occur.  Therefore, if an experimental sets equal to 0.01 which is statistically rare, then we can interpret this mathematically as:

p(event) = 0.01 = 1/100

where p(event) is the probability of the event.

In addition, statistically, null hypothesis signifies no major difference between the specified parameters, and any obvious difference that might occur as a result of experimental error. Thus, it can be concluded that the event is occurring one or fewer times in 100 times if the null hypothesis is true.

3 0
3 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
Really need an answer to this please help.
Alex

Answer:

a

Step-by-step explanation:

\sqrt[4]{144a^{12}b^{3}} = \sqrt[4]{4^{2}*3^{2}a^{12}b^{3}}=\\= \sqrt[4]{2^{4}*3^{2}a^{12}b^{3}}=2a^{3}\sqrt[4]{3^{2}b^{3}} =\\}=2a^{3}\sqrt[4]{9b^{3}}

7 0
3 years ago
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