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Amiraneli [1.4K]
3 years ago
12

Which numbers between 10 and 99 have digits that differ by 4 and add to 6

Mathematics
1 answer:
madreJ [45]3 years ago
8 0

Answer:

A

Step-by-step explanation:

Correct option is A)

Let a 3-digit number ABC

Case (1)

If A=7, then B  and  C can be any digit but not 7

so, total 9 digits (0,1,2,3,4,5,6,8,9) can replace B and C

Then total such combinations =1(9)(9)=81

Case (2)

If B=7  then A can be any digit but not  0 and 7, because if A=0 it will be a 2-digit number.

And C can be any digit but not 7

Then total combinations =(8)(1)(9)=72

Case (3)

If C=7  then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.

And B can be any digit but not 7

Then total combinations =(8)(9)(1)=72

So, total numbers between 99 and 1000 which have exactly one of their digits as 7 are =81+72+72=225

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Which values of x and y would make the following expression represent a real number? (4+5i)(x+yi)
algol [13]

Answer:

So x=4 and y=-5 would work.

Step-by-step explanation:

When you multiply complex conjugates, you get a real result.

Example:

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I replace i^2 with -1 since i^2=-1.

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4-5i

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Hi there!

c=74

<u><em>Equation:</em></u>

26+c=100

c+26=100

100-c=26

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100=26+c

<u><em>1.Find the meaning:</em></u>

so 26 contacts and then the total would be 100 which would be 26+c=100...

<em><u>2.Subtract the two terms:</u></em>

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