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3 years ago

Find the 8th term of the geometric sequence, given the first term and common ratio. a1=6 r=-1/3

Mathematics

1 answer:

Damm [24]3 years ago

8 0

The rule of a geometric sequence:

$a_n=a_1r^{n-1}$

We have:

$a_1=6;\ r=-\dfrac{1}{3}$

Find $a_8=?\to n=8$

substitute:

$a_8=6\cdot\left(-\dfrac{1}{3}\right)^{8-1}=6\cdot\left(-\dfrac{1}{3}\right)^7=6\cdot\left(-\dfrac{1}{2187}\right)=-\dfrac{2}{729}$

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Hope this helps, sorry if this is wrong! Have a great day! <)

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List the steps that should be used to solve the equation f(x) = g(x) by graphing when f(x) = (1/2)² - 3 and g(x) = 3x + 5?

jolli1 [7]

value of x is: $x=\frac{-31}{12}$

Step-by-step explanation:

We need to solve the equation f(x)=g(x)

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$f(x)=g(x)\\(\frac{1}{2})^2-3=3x + 5\\\frac{1}{4}-3=3x + 5\\Subtarct\,\,5\,\,on\,\,both\,\,sides:\\\frac{1}{4}-3-5=3x + 5-5\\\frac{1}{4}-8=3x\\Taking\,\,LCm\,\,on\,\,left\,\,side\\\frac{1-8*4}{4}=3x\\\frac{-31}{4}=3x\\Divide\,\,both\,\,sides\,\,by\,\,3\\x=\frac{-31}{4*3}\\x=\frac{-31}{12}$