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2 years ago

What is the recursive formula, if the sequence is 4, 7, 10, 13, ...?

Mathematics

2 answers:

elena-14-01-66 [18.8K]2 years ago

6 0

Answer:

4 + 3(n-1)

Step-by-step explanation:

Sequences start from n=1,2,3....

In this case

4 + 3(1-1)= 4

4 + 3(2-1) =7

And continue the pattern

tensa zangetsu [6.8K]2 years ago

4 0

Answer:

Step-by-step explanation:

In the arithmetic sequence $t_1$, $t_2$, $t_3$, ..., $t_n$, $t_1=23$ and $t_n= t_{n-1} - 3$ for each n > 1. What is the value of n when $t_n = -4$?

A. -1

B. 7

C. 10

D. 14

E. 20

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_________________

Best Regards,

MGMAT 1 --> 530

MGMAT 2--> 640

MGMAT 3 ---> 610

GMAT ==> 730

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Help me please I have no clue

Lynna [10]

Answer:

Step-by-step explanation:

You see that there’re angles with 30 and 90 degrees that means that -> 180-(90+30)=60 (measurement of the third angle)

Here you’ll need to use the 30,60,90 degree triangle rule:

The opposite side of 30 degree will be y

30->y

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2 years ago

The ratio of pink marbles to white marbles is 1:5 what percent of bag is pink marbles

sashaice [31]

1:5=1/5
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20/100= 20%
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2 years ago

The orange grove produced 892 crates of oranges. Each train car holds 112 crates . What is the minimum number of train cars they

pickupchik [31]

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5 0

3 years ago

Read 2 more answers

Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Sedaia [141]

$\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ 
\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ 
\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation 
becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} 
\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} 
\end{array}$

$\large\begin{array}{l} \textsf{Using 
the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ 
\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ 
\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ 
\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ 
\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot
 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}$

$\large\begin{array}{l}
 \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ 
\mathsf{\Delta=(4.8)^2}\\\\\\ 
\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ 
\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ 
\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 
2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} 
\end{array}$

$\large\begin{array}{l} \begin{array}{rcl} 
\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ 
\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} 
\end{array}$

$\large\begin{array}{l} \textsf{Both 
are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ 
\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or 
}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse 
tangent function:}\\\\ \begin{array}{rcl} 
\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ 
\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ 
\textsf{where k in an integer.} \end{array}$

__________

$\large\begin{array}{l}
 \textsf{Now, restrict x values to the interval 
}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ 
\begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} 
\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{
 is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 
4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} 
\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} 
\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}$

$\large\begin{array}{l}
 \textsf{Solution set:}\\\\ 
\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}
 \end{array}$

<span>If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2071152</span>

$\large\textsf{I hope it helps.}$

Tags: <em>trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function</em>

6 0

2 years ago

Eli was using a recipe for a large smoothie that required three and three–thirds bananas. Instead of making one smoothie, Eli de

notka56 [123]

Answer:

three small smoothies require one and one-third bananas.

Step-by-step explanation:

Assuming that the three small smoothies are equal, and their sum equals the one large smoothie;

Number of bananas required for one large smoothie = three and three-third

three-thirds mean that out of three parts, three parts were selected. Therefore, three-thirds is the same thing as a whole or 1

∴ three and three-third $= 3 + \frac{3}{3} = 3 + 1 = 4$

Therefore, the one large smoothie required 4 bananas.

Large smoothie = 4 bananas

three small smoothies $= \frac{4}{3} = 1 \frac{1}{3}$

Hence, three small smoothies require one and one-third bananas.

6 0

2 years ago