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Papessa [141]
2 years ago
13

Which values of k are solutions to the inequality StartAbsoluteValue negative k minus 2 EndAbsoluteValue less-than 18? Check all

that apply.

Mathematics
2 answers:
Stella [2.4K]2 years ago
6 0

Answer:

You could just replace all the given possible values of k in the inequality and see which ones are solutions, but let's solve this in a more interesting way:

First, remember how the absolute value works:

IxI = x if x ≥ 0

IxI = -x if x ≤ 0

Then if we have something like:

IxI < B

We can rewrite this as

-B < x < B

Now let's answer the question, here we have the inequality:

I-k -2I < 18

Then we can rewrite this as:

-18 < (-k - 2) < 18

Now let's isolate k:

first, we can add 2 in the 3 parts of the inequality:

-18 + 2 < -k - 2 + 2 < 18 + 2

-16 < -k < 20

Now we can multiply all sides by -1, remember that this also changes the direction of the signs, then:

-1*-16 > -1*-k > -1*20

16 > k > -20

Then k can be any value between these two limits.

So the correct options (from the given ones) are:

k = -16

k = -8

k = 0

ddd [48]2 years ago
4 0

Answer:

C, D, E

Step-by-step explanation:

Edge 2021

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liberstina [14]
The 15th term will be 71. Why? Well, see below for an explanation!

By subtracting all of these numbers by the term that comes prior to them, we will find that all of them result in 5. Because of this, we know that each time the term increases, 5 is being added to the numbers. Additionally, I noticed that all of the numbers in this arithmetic sequence only end in a 1 or a 6. Because of this, we can apply the same principle when adding 5 each time:

First term: 1
Second term: 6
Third term: 11
Fourth term: 16
Fifth term: 21
Sixth term: 26
Seventh term: 31
Eighth term: 36
Ninth term: 41
Tenth term: 46
Eleventh term: 51
Twelfth term: 56
Thirteenth term: 61
Fourteenth term: 66
Fifteenth term: 71

By adding 5 each time and keeping in mind that the digits all end in only 1 or 6, we will find that the fifteenth term results in 71. Therefore, the 15th term is 71.

Your final answer: The 15th term of this arithmetic sequence comes down to be 71. If you need extra help, let me know and I will gladly assist you.
4 0
2 years ago
The hypotenuse of a right triangle is 1m longer than twice one of the other two sides. The third side of the triangle is 15m. Fi
Dimas [21]

Let the length of unknown side be x

So, hypotenuse = 2x + 1

By pythagorean theorem ie , a² = b² + c²

(2x + 1)² = x² + 15²

4x² + 1 + 4x = x² + 225

3x² + 4x -224

(3x +28)(x-8)

x = -9.33 , 8

Since, length can't be negative

x = 8

∴Unknown side length = 8 m

Length of hypotenuse = 2 x 8 + 1 = 17 m

Hope it helps now.

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Step-by-step explanation:

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Step-by-step explanation:

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