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Ratling [72]
3 years ago
5

The graph of g(x)=(0.5)x+4 is shown. Which equation is an asymptote of this function? y = 0 y = 4 x = 4 x = 0

Mathematics
1 answer:
Alexandra [31]3 years ago
3 0

we are given

g(x)=(0.5)^x+4

we can see that

there is no value of x for which g(x) is not defined

so, no vertical asymptote exists

now, we will find horizontal asymptote

\lim_{x \to \infty} g(x)= \lim_{x \to \infty}((0.5)^x+4 )

\lim_{x \to \infty} g(x)= (\lim_{x \to \infty} (0.5)^x+\lim_{x \to \infty} 4 )

\lim_{x \to \infty} g(x)= 0+4

so, we get

horizontal asymptote as

y= 4............Answer

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Find the 12th term of the geometric sequence 5, -25, 125, ...5,−25,125,...
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Answer:

  • a_{12}=-244140625

Step-by-step explanation:

Considering the geometric sequence

5,-25,\:125,\:...

a_1=5

As the common ratio 'r' between consecutive terms is constant.

\mathrm{Compute\:the\:ratios\:of\:all\:the\:adjacent\:terms}:\quad \:r=\frac{a_{n+1}}{a_n}

r=\frac{-25}{5}=-5

r=\frac{125}{-25}=-5

The general term of a geometric sequence is given by the formula:  

a_n=a_1\cdot \:r^{n-1}

where a_1 is the initial term and r the common ratio.

Putting n = 12 , r = -5 and a_1=5 in the general term of a geometric sequence to determine the 12th term of the sequence.

a_n=a_1\cdot \:r^{n-1}

a_n=5\left(-5\right)^{n-1}

a_{12}=5\left(-5\right)^{12-1}

      =5\left(-5^{11}\right)

\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a

       =-5\cdot \:5^{11}

\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}

        =-5^{1+11}     ∵ 5\cdot \:5^{11}=\:5^{1+11}

        =-244140625

Therefore,

  • a_{12}=-244140625
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Step-by-step explanation:

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Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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Step-by-step explanation:

<u>Step 1: Define</u>

<u />\frac{-5 \pm \sqrt{5^2-4(3)(1)} }{2(3)}<u />

<u />

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