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3 years ago

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On ITV's Homework Hotline, it's time for "Rudolf's Rules." If you guess what rule Rudolf is using you get a prize. Someone puts

golf balls into a miniature-golf display called Rudolf, then Rudolf sends back a certain number of golf balls. Today Rudolf gets 8 balls and sends back 6. Next Rudolf gets 15 and sends back 27. When he gets 25, 57 are returned. If n represents the number of balls initially given to Rudolf, write an expression to represent how many balls he sent back.

1 answer:

oee [108]3 years ago

7 0

Answer:

y = 3x -18 ( where y = balls sent back by Rudolf; x= balls given to Rudolf)

Step-by-step explanation:

x (Balls given) y( Balls sent back)

8 6 Δy/Δx = (27-6)/(15-8) = 3

15 27

25 57 Δy/Δx = (57-27)/(25-15) = 3

Observing the number sequence shows constant gradient and as such we can use equation of a straight line to relate the two variables.

Using the equation of a straight line we have:

(y-y1)/(x-x1) = (y2-y1)/(x2-x1)

y-6/x-8 = (27-6)/(15-8)

y-6/x-8 = 3

y-6 = 3(x-8)

y-6 = 3x - 24

y = 3x-18

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3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

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Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
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$a \cdot r^2$ ,
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First equation:

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Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

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Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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