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2 years ago

Can someone help with this qs pls? 4 − (−4)^−3

Mathematics

1 answer:

11Alexandr11 [23.1K]2 years ago

4 0

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

Solve ⇨ 4 - (- 4)^-3

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$\sf \: 4 - ( - 4 ) ^ { - 3 }$

Calculate -4 to the power of -3 and get -1/64.

$\dashrightarrow \sf \: ( - 4) ^{3 } \\ = \sf-\frac{1}{4 ^{3} } \\ = \sf - \frac{ 1}{4} \times - \frac{ 1}{4} \times - \frac{ 1}{ 4} \\ = \sf \: - \frac{ 1}{64}$

Come back to the equation & substitute -4³ as -1/64.

$\sf \: 4-\left(-\frac{1}{64}\right) \\$

The opposite of -1/64 is +1/64.

$\sf \: 4+\frac{1}{64} \\$

Add 4 and 1/64 to get 257/64 (take 64 as the LCM).

$\boxed{ \boxed{ \bf\frac{257}{64} = 4.01}}$

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Answer:

Range for third side is

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2 years ago

Find 3rd and 5th form by using nth term formula tn=a+(n-1)d when tn=a+(n-1)d when a=2 and d=3.

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3 years ago

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Which of the following is NOT a linear factor of the polynomial function?

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Answer:

Among the four choices, $(x + 5)$ is the only one that is not a linear factor of this polynomial function.

Step-by-step explanation:

Let $a$ denote some constant. A linear factor of the form $(x - a)$ is a factor of a polynomial $f(x)$ if and only if $f(a) = 0$ (that is: replacing all $x$ in the polynomial $f(x) \!$ with the constant $a\!$ would give this polynomial a value of $0$ .)

For example, in the second linear factor $(x - 2)$ , the value of the constant is $a = 2$ . Verify that the value of $f(2)$ is indeed $0$ . (In other words, replacing all $x$ in the polynomial $f(x) \!$ with the constant $2$ should give this polynomial a value of $0\!$ .)

$\begin{aligned}f(2) &= 2^3 - 5\times 2^2 - 4 \times 2 + 20 \\ &= 8 - 20 - 8 + 20 \\ &= 0 \end{aligned}$ .

Hence, $(x - 2)$ is indeed a linear factor of polynomial $f(x)$ .

Similarly, it could be verified that $(x - 5)$ and $(x + 2)$ are also linear factors of this polynomial function.

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Calculate the value of $f(5)$ .

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$f(5) \ne 0$ implies that $(x - (-5))$ (which is equivalent to $(x + 5)$ ) isn't a linear factor of this polynomial function.

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