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3 years ago

From the 1,2,3,4,5 which values make the inequality n+1 > 4 true?

Mathematics

2 answers:

shusha [124]3 years ago

8 0

Answers: {4, 5}

========================================================

Explanation:

We have some number n and we're adding 1 to it to get n+1.

This n+1 is larger than 4. We want to find which values make n+1 > 4 true.

To find out, undo the operation "add 1" by subtracting 1 from both sides

n+1 > 4

n+1-1 > 4-1

n+0 > 3

n > 3

Anything larger than 3 will work. So that means {4,5} is the solution set from the original list of values.

For instance, if we tried n = 5, then

n+1 > 4

5+1 > 4 ... replace n with 5

6 > 4 ... true statement

But if we tried n = 1, then,

n+1 > 4

1+1 > 4

2 > 4 ... false; this shows n = 1 isn't in the solution set.

ELEN [110]3 years ago

4 0

Answer:

yes it is

Step-by-step explanation:

hope I helped;)

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Example:

If $f(x)=\sin x$ , then we have $\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x$ , and so $\sin x$ is periodic with period $2\pi$ .

It gets a bit more complicated for a function like yours. We're looking for $k$ such that

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It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when $k$ is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.

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$f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))$

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