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3 years ago

Solve the equation s/r -2=t/r solve for r

Mathematics

1 answer:

vivado [14]3 years ago

8 0

Answer: have a good day :)

$r=\frac{s}{2}-\frac{t}{2}$

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Evaluate piecewise functions <br> Pls answer

Andrew [12]

Answer:

-23

Step-by-step explanation:

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= -20 - 3

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3 years ago

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3 years ago

Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -6 and 162, respectively.

Vesna [10]

Answer:

b. an = 2 • (-3)^(n - 1)

Step-by-step explanation:

Before we test a solution or two, we can easily discard most of them.

We see the values alternate of signs (-5 for the 2nd term and +162 for the 5th term)... so the progression factor has to be negative (in order to alternate sign). <u>That already excludes answers A and C.</u>

Normally, a geometric progression has the (n-1) exponent, not (n+1), so our chances seem to be better with B.

We can test both D and B with n = 2, to obtain -6

Let's test answer D before:

$a_{2} = 2 * (-3)^{2+1} = 2 * (-3)^{3} = 2 * -27 = -54$

The result is -54, not -6... so it's not the right result.

Let's test answer B then:

$a_{2} = 2 * (-3)^{2-1} = 2 * (-3)^{1} = 2 * -3 = -6$

Right! Let's verify with n=5 to get 162:

$a_{5} = 2 * (-3)^{5-1} = 2 * (-3)^{4} = 2 * 81 = 162$

Confirmed, answer is B. an = 2 • (-3)^(n - 1)

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3 years ago

Another question I need help with. I've got the part where it says to rewrite as a radical expression on my paper, but I'm havin

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$\bf a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} 
\qquad \qquad
\sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}}\\\\
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32=2\cdot 2\cdot 2\cdot 2\cdot 2\\
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2^3\implies 8$

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4 years ago

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Answer:

9 hours

Step-by-step explanation:

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4 years ago

Solve the equation s/r -2=t/r solve for r​

Solve the equation s/r -2=t/r solve for r