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ExtremeBDS [4]
2 years ago
9

(-5z)^3 using only positove exponents

Mathematics
1 answer:
Elena L [17]2 years ago
8 0

Answer:

Shown below

Step-by-step explanation:

First allocate the power of three to every number

-5^{3}  * z^{3} = -125 * z^{3}

not sure what you mean by positive exponents, the exponents already are positive. Maybe you mean positive exponents and integers?

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How to solve 76×327 step by step
hram777 [196]

Here's a method that takes advantage of the values of these particular numbers.

... 76 = 80 - 4

so

... 76 × 327 = (80 -4)×327

... = 80×327 - 4×327

Repeated doubling will give us values that are 2, 4, and 8 times 327.

... 2×327 = 327+327 = 654

... 2×654 = 4×327 = 654+654 = 1308 . . . . we'll use this later

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We want 80×327, so we can add a zero to the end of this last:

... 80×327 = 26160

Now, we can subtract 4×327 to get 76×327

... 80×327 - 4×327 = 26160 -1308 = 24852 = 76×327

_____

More conventionally, you would multiply every digit of one number by every digit of the other and add the products according to their respective place values.

327 × 076 = (3×0)×10000 + (3×7 +2×0)×1000 +(3×6 +7×0 +2×7)×100 +(2×6 +7×7)×10 +(7×6)×1

... = 0 +21,000 +3,200 + 610 +42

... = 24,852

Note the pattern of partial products here. This is a method taught to/by practitioners of Vedic mathematics, and can be done in your head. At most, you would write down the partial product sums 21, 32, 61, and 42 to keep from having to carry more than one number in your head at a time.

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