Ill need asap please

Mathematics

2 answers:

murzikaleks [220]1 year ago

5 0

D. 37.8204

A negative absolute value is always positive.

Firlakuza [10]1 year ago

4 0

Answer:

d) 37.8204

Step-by-step explanation:

Given that,

→ |-18.531| + |-19.2894|

Let's solve the problem,

→ |-18.531| + |-19.2894|

→ 18.531 + 19.2894

→ 37.8204

Thus, option (d) is correct.

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Find the derivative of the following functions: a. f(x) = (x^3 + 5)^1/4 - 15e^x^3 b. f(x) = (x - 3)^2 (x - 5)/(x - 4)^2(x^2 + 3)

Darya [45]

Answer:

Step-by-step explanation:

Given function is

(a)F(x)= $\left ( x^{3}+5\right )^{0.25}-15e^{x^{3}}$

$F^{'}\left ( x\right )$ = $0.25\left ( x^{3}+5\right )^{-0.75}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}-15e^{x^{3}}\frac{\mathrm{d} x^{3}}{\mathrm{d} x}$

$F^{'}\left ( x\right )=0.25\left ( x^{3}+5\right )^{-0.75}\times 3x^{2}-15e^{x^{3}}\times \left ( 3x^{2}\right )$

(b)F(x)= $\frac{\left ( x-3\right )^2\left ( x-5\right )}{\left ( x-4\right )^2\left ( x^{2}+3\right )^5}$

$F^{'}\left ( x\right )$ = $\frac{\left [ 2\left ( x-3\right )\right \left ( x-5\right )+\left ( x-3\right )^2]\left [ \left ( x-4\right )^2\left ( x^2+3\right )^5\right ]-\left [ 2\left ( x-4\right )^{3}\left ( x^2+3\right )^5+5\left ( x^2+3\right )^4\left ( 2x\right )\left ( x-4\right )^2\right ]\left [ \left ( x-3\right )^2\left ( x-5\right )\right ]}{\left [\left ( x-4\right )^2\left ( x^2+3\right )^5\right ]^2}$