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ruslelena [56]
3 years ago
15

For what value of a does 9 = (StartFraction 1 Over 27 EndFraction) Superscript a + 3?

Mathematics
1 answer:
vodka [1.7K]3 years ago
7 0

Answer:

A

Step-by-step explanation:

9 = (1/27)^(a+3)

3² = (3^-3)^(a+3)

3² = 3^(-3a-9)

2 = -3a-9

-3a = 11

a = -11/3

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In 1998, Cathy's age is equal to the sum of the four digits in the year of her birthday, then how old was Cathy in 1998?
ycow [4]

Answer:

<em>Cathy was born in 1980 and she was 18 years old in 1998</em>

Step-by-step explanation:

<u>Equations</u>

This is a special type of equations where all the unknowns must be integers and limited to a range [0,9] because they are the digits of a number.

Let's say Cathy was born in the year x formed by the ordered digits abcd. A number expressed by its digits can be calculated as

x=1000a+100b+10c+d

In 1998, Cathy's age was

1998-(1000a+100b+10c+d)

And it must be equal to the sum of the four digits

1998-(1000a+100b+10c+d)=a+b+c+d

Rearranging

1998=1001a+101b+11c+2d

We are sure a=1, b=9 because Cathy's age is limited to having been born in the same century and millennium. Thus

1998=1001+909+11c+2d

Operating

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If now we try some values for c we notice there is only one possible valid combination, since c and d must be integers in the range [0,9]

c=8, d=0

Thus, Cathy was born in 1980 and she was 18 years old in 1998. Note that 1+9+8+0=18

5 0
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