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2 years ago

13

Evacuate the following using suitable identities (102)^3

1 answer:

snow_lady [41]2 years ago

5 0

The value of $(102)^3$ exists 1061208.

<h3>How to estimate the value of $(102)^3$ ?</h3>

Let us rewrite $(102)^3$ as $(100+2)^3$

Now utilizing the identity $(a+b)^3=a^3+b^3+3ab(a+b)$ , we get

a = 100 and b = 2 then substitute the values of a and b then

$(100+2)^3=100^3+2^3+[(3\times100\times2)(100+2)]$

= 1000000 + 8 + (600 × 102)

= 1000000 + 8 + 61200

= 1061208

Hence, $(102)^3=1061208$

Therefore, the value of $(102)^3$ exists 1061208.

To learn more about cubic polynomial equation refer to:

brainly.com/question/28181089

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2 years ago

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riadik2000 [5.3K]

<h3>Answer:</h3>

a) l = 5 + 2w

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<h3>Explanation:</h3>

a) Using w for width, twice the width is 2w, and 5 more than that is 5+2w. This is your expression for l:

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b) The perimeter is the sum of the lengths of all sides, so is the sum of twice the length and twice the width.

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