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IrinaVladis [17]
3 years ago
13

What is the identity of 8z^3+27

Mathematics
1 answer:
Alik [6]3 years ago
8 0

8 is a cube: 2³

27 is a cube: 3³

The expression can be identified as "the sum of two cubes": (2z)³ + 3³.

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the tens digit of a two digit number is 5 greater the units digit. If you subtract double the reversed number from it, the resul
likoan [24]

Given:

The tens digit of a two digit number is 5 greater the units digit.

If you subtract double the reversed number from it, the result is a fourth of the original number.

To find:

The original number.

Solution:

Let n be the two digit number and x be the unit digit. Then tens digit is (x+5) and the original number is:

n=(x+5)\times 10+x\times 1

n=10x+50+x

n=11x+50

Reversed number is:

x\times 10+(x+5)\times 1=10x+x+5

x\times 10+(x+5)\times 1=11x+5

If you subtract double the reversed number from it, the result is a fourth of the original number.

11x+50-2(11x+5)=\dfrac{1}{4}(11x+50)

11x+50-22x-10=\dfrac{1}{4}(11x+50)

40-11x=\dfrac{1}{4}(11x+50)

Multiply both sides by 4.

160-44x=11x+50

160-50=11x+44x

110=55x

Divide both sides by 55.

\dfrac{110}{55}=x

2=x

The unit digit is 2. So, the tens digit is 2+5=7.

Therefore, the original number is 72.

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