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2 years ago

Find the smallest value of k such that 16k is a perfect cube.

Mathematics

2 answers:

Leona [35]2 years ago

4 0

Answer:

k = 4

Step-by-step explanation:

Knowing that 64 is a perfect cube and 64 = 16 × 4, then

16k ← with k = 4 is a perfect cube, that is

$\sqrt[3]{64}$ = 4

IrinaK [193]2 years ago

3 0

Answer:

Step-by-step explanation:

The smallest perfect cube would be 0 since it cannot be negative and zero raised to the third power would still be 0

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olganol [36]

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2 years ago

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jonny [76]

Answer:

Step-by-step explanation:

f(x) = 2ˣ + 1

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2 years ago

Based on the polynomial remainder theorem, what is the value of the function when x=−6 ?

Vaselesa [24]

ANSWER

$f( - 6) = 10$

EXPLANATION

The given function is
$f(x) = {x}^{4} + 8 {x}^{3} + 10 {x}^{2} - 7x + 40$

According to the remainder theorem,

$f( - 6) = R$
where R is the remainder when
$f(x)$

is divided by
$x + 6$

We therefore substitute the value of x and evaluate to obtain,

$f( - 6) = { (- 6)}^{4} + 8 {( - 6)}^{3} + 10 { (- 6)}^{2} - 7( - 6)+ 40$

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$f( - 6) = 1296 - 1728 + 360 + 42+ 40$

This will evaluate to,

$f( - 6) = 10$

Hence the value of the function when
$x = - 6$

is
$10$

According to the remainder theorem,
$10$
is the remainder when

$f(x) = {x}^{4} + 8 {x}^{3} + 10 {x}^{2} - 7x + 40$
is divided by
$x + 6$

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2 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Find the smallest value of k such that 16k is a perfect cube.​

Find the smallest value of k such that 16k is a perfect cube.