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3 years ago

IN YOUR OWN WORDS How is the cube root of a number different from the square root of a number?

Mathematics

2 answers:

ella [17]3 years ago

8 0

Answer: the cube root would be finding 3 multiples square root would be two

kiruha [24]3 years ago

7 0

Answer:

Just as the square root is a number that, when squared, gives the radicand, the cube root is a number that, when cubed, gives the radicand. Cubing a number is the same as taking it to the third power: 23 is 2 cubed, so the cube root of 23 is 2.

Step-by-step explanation:

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3 years ago

Hi, Could anyone help me with my homework

tatiyna

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$\hookrightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

solving steps:

$\rightarrow \sf (x + 4)^6$

$\bold{rewrite \ the \ following}$

$\rightarrow \sf (x + 4)^2 (x + 4)^2 (x + 4)^2$

$\bold {formula \ used : \sf (x+a)^2 = (x^2 + 2xa + a^2)}$

$\rightarrow \sf (x^2 + 8x+16) (x^2 + 8x+16) (x^2 + 8x+16)$

$\bold{simplify \ by \ removing \ parenthesis}$

$\rightarrow \sf (x^4 +8x^3 + 16x^2 + 8x^3 +64x^2 + 128x+16x^2+128x+256 ) (x^2 + 8x+16)$

$\bold{basic \ addition \ of \ integers }$

$\rightarrow \sf (x^4+16x^3+96x^2+256x+256) (x^2 + 8x+16)$

$\bold{remove \ parenthesis}$

$\rightarrow \sf (x^6 + 16x^5 + 96x^4 + 256x^3 + 256x^2 + 8x^6 + 128x^4 + 768x^3 + 2048x^2 + 2048 + 16x^4 + 256x^3 + 1536x^2 + 4096x + 4096)$

$\bold {final \ answer:}$

$\rightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

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2 years ago

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SpyIntel [72]

Answer:

No, none of the number need to be 48 for the mean to be 48. To get a mean, you add up all the number and divide it by the amount of numbers.

Example:

the mean of 10, 79, 42, 88, 19, and 50 is 48, but the actual number 48 was not part of the set.

10 + 79 + 42 + 88 + 19 + 50 = 288

288 ÷ 6 = 48

6 0

3 years ago

IN YOUR OWN WORDS How is the cube root of a number different from the square root of a number?​

IN YOUR OWN WORDS How is the cube root of a number different from the square root of a number?