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kumpel [21]
2 years ago
13

How many positive 3-digit numbers have property that the first digit is at least three times the second digit?

Mathematics
1 answer:
ss7ja [257]2 years ago
3 0

The number of positive 3-digit numbers have a property that the first digit is at least three times the second digit is 285.

<h3>How to get the number?</h3>

Since restriction is set to the first digit, all you have to think is about the last two digits e.g if the first digit is one, the last two digits can only be zeros.

Thus, by the principle of multiplication, 1×1=1 way. If the first digit is 2, then the last digits can be either 1 or 0. You have 2×2=4 ways

For first digit 3, the last two digits can be 0,1 or 2. 3x3=9, there is a succinct pattern. Therefore, 1+4+9+16+25+36+49+64+81=285.

Learn more about numbers on:

brainly.com/question/251701

#SPJ1

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Xelga [282]

To find the pre image you need to back track on the image. To get to the image you used (x-6,y+8). Now you need to use the exact opposite to get back to the pre image. For this you would change the signs to look like (x+6,y-8). Now we just apply this to (-4,1).

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(2,-7) should be the pre image point.

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mestny [16]

Answer:

(x^2-3^2)(x+3)

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Step-by-step explanation:

Given

(x^2-9)(x+3)

Required

Select 3 equivalent expressions

(x^2-9)(x+3)

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Take x^2 - 3^2 as difference of two squares

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Solving further:

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(x-3)(x+3)^2   -- This is another equivalent expression:

Hence, the equivalent expressions are:

(x^2-3^2)(x+3)

(x-3)(x+3)(x+3)

(x-3)(x+3)^2

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Step-by-step explanation:

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