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2 years ago

The nth term of a sequence is given by 3n² + 11 Calculate the difference between the 6th term and the 9th term of the sequence.

Mathematics

1 answer:

katovenus [111]2 years ago

4 0

The difference between the 6th term and the 9th term of the sequence is 135

<h3>How to determine the difference</h3>

Given that the nth term is;

3n² + 11

For the 6th term, the value of n is 6

Let's solve for the 6th term

= 3( 6)^2 + 11

= 3 × 36 + 11

= 108 + 11

= 119

For the 9th term, n = 9

= 3 (9)^2 + 11

= 3( 81) + 11

= 243 + 11

= 254

The difference between the 6th and 9th term

= 254 - 119

= 135

Thus, the difference between the 6th term and the 9th term of the sequence is 135

Learn more about algebraic expressions here:

brainly.com/question/4344214

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3 years ago

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d1i1m1o1n [39]

<u>Given</u>:

The given expression to find the nth term of the sequence is $d(n)=d(n-1) \cdot (-5)$

The first term of the sequence is $d(1)=8$

We need to determine the third term of the sequence.

<u>Second term:</u>

The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

Thus, we have;

$d(2)=d(2-1) \cdot (-5)$

$d(2)=d(1) \cdot (-5)$

$d(2)=8 \cdot (-5)$

$d(2)=-40$

Thus, the second term of the sequence is -40.

<u>Third term:</u>

The third term of the sequence can be determined by substituting n = 3 in the nth term of the sequence.

Thus, we have;

$d(3)=d(3-1) \cdot (-5)$

$d(3)=-40 \cdot (-5)$

$d(3)=120$

Thus, the third term of the sequence is 120.

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3 years ago

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Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

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Also notice, the "c" distance from the center to either foci, is just 5 units.

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

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

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3 years ago

The nth term of a sequence is given by 3n² + 11 Calculate the difference between the 6th term and the 9th term of the sequence.​

The nth term of a sequence is given by 3n² + 11 Calculate the difference between the 6th term and the 9th term of the sequence.