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blsea [12.9K]
1 year ago
7

Which of the statements are true about the function f given by f(x) = 100-e? Select all that apply.

Mathematics
1 answer:
tatiyna1 year ago
6 0

From the above function, it is clear that the value of f is never 0. Hence the statement that is true is (Option E), See explanation of same below.

<h3>What is the explanation for the above function?</h3>

Note that the function is related to Euler's number which is depicted as:
e ≈ 2.7182. The function is given as:

f(x) = 100 * e^{-x}

Assuming x = -2, we'd have:

100 * 2.7182^{-2}

= 271.82^{-2}

= 0.00001353354

Hence, even when x tends < 0 the function f(x) thus, is never 0. See the attached graph for confirmation.

Learn more about functions at:

brainly.com/question/25638609

#SPJ1

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Bess [88]

Answer:

<u><em>F(x)= 5*[\frac{x^{3} }{3} + (a*b)*\frac{x^{2} }{2} + a*b*x + C.</em></u>

Step-by-step explanation:

<u><em>First step we aplicate distributive property to the function.</em></u>

<u><em>5*(x+a)*(x+b)= 5*[x^{2}+x*b+a*x+a*b]</em></u>

<u><em>5*[x^{2}+x*(b+a)+a*b]= f(x), where a, b are constant and a≠b</em></u>

<u><em>integrating we find ⇒∫f(x)*dx= F(x) + C, where C= integration´s constant</em></u>

<u><em>∫^5*[x^{2}+x*(a+b)+a*b]*dx, apply integral´s property</em></u>

<u><em>5*[∫x^{2}dx+∫(a*b)*x*dx + ∫a*b*dx], resolving the integrals </em></u>

<u><em>5*[\frac{x^{3} }{3} + (a*b)*\frac{x^{2} }{2} + a*b*x</em></u>

<u><em>Finally we can write the function F(x)</em></u>

<u><em>F(x)= 5*[\frac{x^{3} }{3} + (a*b)*\frac{x^{2} }{2} + a*b*x ]+ C.</em></u>

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