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VLD [36.1K]
3 years ago
14

Find the number set which satisfies each of the problems. If 20 is added to the number, the absolute value of the result is 6.

Mathematics
1 answer:
victus00 [196]3 years ago
7 0

Answer: {-14,-26}

Step-by-step explanation:

Let x be the number.

Given statements:  If 20 is added to the number, the absolute value of the result is 6.

i.e. |x+20| =6

Since |y|=c \Rightarrow y= c \text{ or } y = -c

So, |x+20| =6\Rightarrow\ x+20=6\text{ or }x+20=-6

\Rightarrow\ x= 6-20\text{ or } x= -6-20\\\\\Rightarrow\ x=-14\text{ or } x=-26

hence, set of values of x = {-14,-26}

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Baking c cookies and dividing them evenly among 4 friends.

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Point A is located at (1, 5), and point M is located at (−1, 6). If point M is the midpoint of segment AB, find the location of
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HELP MEEEEEEEEEEEE PLZ
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The first 25 prime numbers (all the prime numbers less than 100) are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

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3 years ago
What is the sum of series is numbers 64&amp;66 ?<br> Using the summation notation
iragen [17]

Answer:

64. 6138

66. 8.078125

Step-by-step explanation:

Let's take it one number at a time.

For number 64:

6(2)^{1-1}=6

6(2)^{2-1}=12

6(2)^{3-1}=24

6(2)^{4-1}=48

6(2)^{5-1}=96

6(2)^{6-1}=192

6(2)^{7-1}=384

6(2)^{8-1}=768

6(2)^{9-1}=1536

6(2)^{10-1}=3072

Sum = 6 + 12 + 24 + 48 + 96 + 192 + 384 + 768 + 1536 + 3072

Sum = 6138

For number 66:

12(-\frac{1}{2})^{0}=12

12(-\frac{1}{2})^{1}=-6

12(-\frac{1}{2})^{2}=3

12(-\frac{1}{2})^{3}=-1.5

12(-\frac{1}{2})^{4}=0.75

12(-\frac{1}{2})^{5}=-0.375

12(-\frac{1}{2})^{6}=0.1875

12(-\frac{1}{2})^{7}=-0.0078125

12(-\frac{1}{2})^{8}=0.046875

12(-\frac{1}{2})^{9}=-0.0234375

Sum = 12 + -6 + 3 + -1.5 + 0.75 + -0.375 + 0.1875 + -0.0078125 + 0.046875 + -0.0234375

or to make things simpler:

Sum = 12 - 6 + 3 - 1.5 + 0.75 - 0.375 + 0.1875 - 0.0078125 + 0.046875 - 0.0234375

Sum = 8.078125

4 0
3 years ago
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