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2 years ago

Find a₁0 given that a7 = 6 and d = 3 in the arithmetic sequence.

Mathematics

1 answer:

levacccp [35]2 years ago

7 0

The tenth term of the arithmetic sequence will be 15. Then the correct option is A.

<h3>What is the arithmetic sequence?</h3>

Let a₁ be the first term and d is the common difference between the terms. Then the nth term will be given as

$\rm a_n = a_1 + (n - 1)d$

We have

a₇ = 6

d = 3

n = 7

Then the first term will be

a₇ = a₁ + (7 – 1)3

6 = a₁ + 18

a₁ = 6 – 18

a₁ = -12

Then the 10th term will be

a₁ = -12

d = 3

n =10

Then we have

a₁₀ = -12 + (10 – 1)3

a₁₀ = -12 + 27

a₁₀ = 15

Then the tenth term of the arithmetic sequence will be 15.

Then the correct option is A.

More about the arithmetic sequence link is given below.

brainly.com/question/15412619

#SPJ1

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A hyperbola centered at the origin has a vertex at (9, 0) and a focus at (−15, 0). Which are the equations of the asymptotes? y

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Well, we know is centered at the origing, thus h,k are just 0,0.

using the provided vertex and focus point, gives us a distance for "a" of 9 and a distance for "c" of 15, check the picture below.

$\bf \textit{hyperbolas, horizontal traverse axis }\\\\
\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2+ b ^2}\\
\textit{asymptotes}\quad 
y= k\pm \cfrac{b}{a}(x- h)
\end{cases}\\\\
-------------------------------$

$\bf \begin{cases}
a=9\\
c=15
\end{cases}\implies c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{15^2-9^2}=b\implies \sqrt{144}=b\implies \boxed{12=b}\\\\
-------------------------------\\\\
asymptotes\qquad y=0\pm\cfrac{12}{9}(x-0)\implies y=\pm\cfrac{4}{3}x$

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