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2 years ago

Evaluate f(t)=-2t2+ 1 for f(-3).

Mathematics

1 answer:

podryga [215]2 years ago

3 0

Answer:

Step-by-step explanation:

First, plug-in f(-3) into t in the function:

$f(-3)=-2(-3) +1$

Second, multiply -2 and -3 to get 6:

$f(-3)=-2(-3)+1$ ⇒ $f(-3)=6+1$

Third, add 6 and 1 to get your answer:

$f(-3)= 6+1$ ⇒ $f(-3)=7$

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16.6

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2 years ago

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Answer:

interval = [17.3948 , 20.6052]

Step-by-step explanation:

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2 years ago

What is the equation of the line that represents the horizontal asymptote of the function f(x)=25,000(1+0.025)^(x)?

posledela

Answer:

The answer is below

Step-by-step explanation:

The horizontal asymptote of a function f(x) is gotten by finding the limit as x ⇒ ∞ or x ⇒ -∞. If the limit gives you a finite value, then your asymptote is at that point.

$\lim_{x \to \infty} f(x)=A\\\\or\\\\ \lim_{x \to -\infty} f(x)=A\\\\where\ A\ is\ a\ finite\ value.\\\\Given\ that \ f(x) =25000(1+0.025)^x\\\\ \lim_{x \to \infty} f(x)= \lim_{x \to \infty} [25000(1+0.025)^x]= \lim_{x \to \infty} [25000(1.025)^x]\\=25000 \lim_{x \to \infty} [(1.025)^x]=25000(\infty)=\infty\\\\ \lim_{x \to -\infty} f(x)= \lim_{x \to -\infty} [25000(1+0.025)^x]= \lim_{x \to -\infty} [25000(1.025)^x]\\=25000 \lim_{x \to -\infty} [(1.025)^x]=25000(0)=0\\\\$

$Since\ \lim_{x \to -\infty} f(x)=0\ is\ a\ finite\ value,hence:\\\\Hence\ the\ horizontal\ asymtotes\ is\ at\ y=0$

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2 years ago

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2 years ago