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WARRIOR [948]
3 years ago
5

How many three digit counting numbers are not multiples of 10?

Mathematics
1 answer:
seraphim [82]3 years ago
6 0

Three digits number go from 100 to 999, and thus there are 999-100+1 = 900 of them.


The multiples of ten are numbers ending with zero, so we must exclude all numbers like xy0.


The first digit, x, range from 1 to 9, while the second digit, y, range from 0 to 9.


So, to generate a multiple of ten between 100 and 999, we have 9 choices for the first digit, 10 choices for the second, while the third digit is fixed to zero.


This means that there are 9 \cdot 10 = 90 multiples of ten between 100 and 999.


Since there are 900 numbers in total, there are 900-90 = 810 numbers between 100 and 999 that are not multiples of ten.

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3 years ago
Help, two of these questions plz
Olenka [21]

Answer:

First Answer: y= 64 and x=64

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4 0
3 years ago
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How many and what type of solutions does the equation have?
Natali [406]
Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0

As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ<u /> = 9 + 72
Δ<u /> = 81

Δ<u />>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2

So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2

A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.

3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.

So 2k^2 = 9 + 3k have two rational solutions (Option B).

Hope this Helps! :)
7 0
3 years ago
Read 2 more answers
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