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3 years ago

Cube roots of 1 are +1 and -1 (true and false)

Mathematics

1 answer:

mario62 [17]3 years ago

7 0

False cube root of 1 is plus 1 only.

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Need help with this.

sweet [91]

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Let t as y² + 5y .

${y}^{2} + 5y = t$

Rewrite the equation :

$t - \frac{36}{t} = 0 \\$

Multiply sides by t

$t \times (t - \frac{36}{t} ) = t \times 0 \\$

${t}^{2} - 36 = 0$

Add sides 36

${t}^{2} - 36 + 36 = 0 + 36$

${t}^{2} = 36$

$t = ± \: 6$

_________________________________

So :

${y}^{2} + 5y = 6$

Subtract sides 6

${y}^{2} + 5y - 6 = 6 - 6$

${y}^{2} + 5y - 6 = 0$

$(y + 6)(y - 1) = 0$

$y = - 6 \: \: or \: \: y = 1$

##############################

${y}^{2} +5y = - 6$

$No \: \: solution$

_________________________________

Thus (( y = 1 )) and (( y = - 6 )) are the roots of the equation .

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3 years ago

Help or ill die so please help!!!!!!!!!!!!!!!!!!!!

alekssr [168]

Answer: the answer is b

Step-by-step explanation:

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Cube roots of 1 are +1 and -1 (true and false)​

Cube roots of 1 are +1 and -1 (true and false)