All categories

3 years ago

What number pattern is generated by the following

Mathematics

1 answer:

jenyasd209 [6]3 years ago

3 0

Answer:

nth term = 1 + n(13/4)

Step-by-step explanation:

3n + n/4 + 1

n = 1, 3(1) + (1)/4 + 1 = 4 + ¼ = 17/4

n = 2, 3(2) + (2)/4 + 1 = 7 + ½ = 30/4

n = 3, 3(3) + (3)/4 + 1 = 10 + ¾ = 43/4

n = 4, 3(4) + (4)/4 + 1 = 14 = 56/4

The difference between consecutive terms is 13/4

At n = 1, 17/4 = 1 + 1(13/4)

At n = 2, 30/4 = 1 + 2(13/4)

nth term = 1 + n(13/4)

You might be interested in

Macys is having a sale on all shoes, if I buy a pair of boots that are $125 but are now 30% off what is the sale price of the bo

natita [175]

$87.5 because 30% of 125 is 37.5 so 125-37.5 is 87.5

8 0

2 years ago

59 and/or 60. Determine if the triangles are similar. If they are similar, find the scale factor from ABC to JKL.

navik [9.2K]

9514 1404 393

Answer:

59) not similar

60)

Step-by-step explanation:

59) The ratios of the side lengths shown cannot be reduced. They are different, so the triangles are not similar.

7:8:12 ≠ 6:7:11

60) The side length ratios both reduce to 8:7:10, so the triangles are similar.

∆ABC ~ ∆JKL

The scale factor is LJ/CA = 25/20 = 5/4. (Multiplying ABC by 5/4 will give JKL.)

8 0

2 years ago

Caine

garri49 [273]

Answer:

well sorry I dont understand you

3 0

2 years ago

What is 420 in 7hours

son4ous [18]

Answer:

Step-by-step explanation:

420 divided by 7

4 0

3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}

\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\qquad \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\qquad \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)
\end{array}$

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago