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leva [86]
3 years ago
10

Find the sum of the geometric sequence 3, 15, 75, 375, … when there are 9 terms and select the correct answer below.

Mathematics
1 answer:
jeka943 years ago
3 0

Answer:

  c.  1,464,843

Step-by-step explanation:

The sum of n terms of a geometric sequence with first term a1 and common ratio r is given by ...

  sn = a1(r^n -1)/(r -1)

Filling in the values a1=3, r=5, n=9, we get ...

  s9 = 3(5^9 -1)/(5 -1) = 1,464,843

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The probability that the bus is late on Monday is 20%, and the probability that the bus is late on Tuesday is 10%. What is the p
Aleks [24]
Add 20% and 10% (30%). Then, because it is two days combined, it would be 30% of 200. Divide each by 2, and there is your answer. 

<span>The probability that the bus will be late on both days is: 15%.</span>
7 0
3 years ago
A drum of oil has a height of 4 feet and a radius of 1.2 feet. the oil cost $22 per cubic feet.
blondinia [14]
Well, first we should calculate just how much volume the drum can carry.
The formula we should use is πr^2(h):
π((1.2 ft)^2)(4 ft)
π(1.44 ft^2)(4 ft)
5.76π ft^3 ≈ 18.1 ft^3
Thus is costs:
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4 0
3 years ago
Read 2 more answers
Pueden responder rápido por favor? tengo hasta las 10 :v
Marina CMI [18]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
9 months ago
Match the missing term with the perfect square trinomial 9X^2+?+25<br> 49<br> 6x<br> 36<br> 30x
Igoryamba
9x^2+t+25=(ax+b)^2

9x^2+t+25=a^2*x^2+2abx+b^2 so

a^2=9 so a=3

b^2=25 so b=5 then

t=2ab using a and b from above

t=2*3*5=30

So the missing term is 30x
4 0
2 years ago
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