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Mathematics

1 answer:

zepelin [54]2 years ago

4 0

Answer:

25%

Step-by-step explanation:

multiply it by 100 over 1 then divide the top by the bottom.

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Explain why (–4x)0 = 1, but –4x0 = –4.

hodyreva [135]

Any number raised to power of 0 is 1.
In (-4x)^0, because of the parenthesis, the whole number is being raided to 0, which gives 1.
But in -4x^0, only x is being raised to 0, thus -4x^0 = -4(1) = -4.

5 0

3 years ago

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

3 years ago

Evaluate this exponential expression (27/8)^ 4/3 A: 9/2 B: 4/3 C: 81/16

valkas [14]

$\bf ~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \left( \cfrac{27}{8} \right)^{\frac{4}{3}}\implies \left( \cfrac{3^3}{2^3} \right)^{\frac{4}{3}}\implies \left( \cfrac{3^{3\cdot \frac{4}{3}}}{2^{3\cdot \frac{4}{3}}} \right)\implies \cfrac{3^4}{2^4}\implies \cfrac{81}{16}$