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Nikitich [7]
3 years ago
13

13)

Mathematics
2 answers:
Alenkasestr [34]3 years ago
6 0

Answer:

22

Step-by-step explanation:

did the test with that question

Anna007 [38]3 years ago
4 0

Answer:

22

Step-by-step explanation:

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The sum of two complex numbers, where the real numbers do not equal zero, results in a sum of 34i. Which statement must be true
nekit [7.7K]

Answer:

It's 4 ;)

Step-by-step explanation:

Real numbers that are opposite would be a way that they would cancel out, leaving the imaginary number components.

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3 years ago
1
Reptile [31]

Moving up would add to the Y value. Moving to the left subtracts from the X value.

The answer would be: (X,Y) — (x-3, y + 5)

4 0
3 years ago
1)10 = z+ 6
Fofino [41]
1. Z=4 10-6
2 y= 6 48/8
3. Q= 13 12+1
6 m=15 11+4
7. 21 19+2
8. S=2 3-1
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8 0
3 years ago
Is 13 a perfect square
ladessa [460]
1^2=1 \\ 2^2=4 \\ 3^2=9 \\ 4^2=16 \\ 5^2=25 \\ 6^2=36 \\ 7^2=49 \\ 8^2=64 \\ 9^2=81 \\ 10^2=100 \\ 11^2=121 \\ 12^2=144 \\ \boxed{13^2=169} \\  14^2=196 \\ 15^2=225 \\
8 0
4 years ago
Find the smallest positive integer n such that the digit sum of n is divisible by 5, and the digit sum of n +1 is also divisible
alisha [4.7K]

Answer:

139,999

Step-by-step explanation:

If the digit sum of n is divisible by 5, the digit sum of n+1 can't physically be divisble by 5, unless we utilise 9's at the end, this way whenever we take a number in the tens (i.e. 19), the n+1 will be 1 off being divisble, so if we take a number in the hundreds, (109, remember it must have as many 9's at the end as possible) the n+1 will be 2 off being divisble, so continuing this into the thousands being three, tenthousands being 4, the hundred thousands will be 5 off (or also divisble by 5). So if we stick a 1 in the beginning (for the lowest value), and fill the last digits with 9's, we by process of elimination realise that the tenthousands digit must be 3 such that the digit sum is divisible by 5, therefore we get 139,999

6 0
2 years ago
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