Measurement and Monetary Arithmetic | The Final Information to NESA’s Maths Reference Sheet

(l=frac{theta}{360}times2pi r)

Space

(A= frac{theta}{360} occasions pi r^2)

(A=frac{h}{2} (a+b) )

(A=2 pi r^2 + 2 pi r h )

(A = 4 pi r^2)

Quantity

(V= frac{1}{3}Ah)

(V= frac{4}{3} pi r^3)

Sequences and sequence

(T_n = a+(n-1)d)

(S_n=frac{n}{2} [2a+(n-1)d] = frac{n}{2}(a+l))

(T_n=ar^{n-1})

(S_n=frac{a(1-r^n)}{1-r}=frac{a(r^n-1)}{r-1}, r≠1)

(S= frac{a}{1-r}, |r|<1 )

(utilizing θ in levels)

start{align*}
l &=frac{115}{360}times2pi occasions 30
&= frac{115}{6} pi
∴ textual content{Size of the minor arc AB} &= frac{115}{6} pi textual content{ cm}
finish{align*}

start{align*}
A &= frac{330}{360} occasions pi occasions {5}^2
&= frac{275}{12} pi
∴ textual content{Space of the most important sector AOB} &= frac{275}{12} pi textual content{ m}^2
finish{align*}

start{align*}
A &= frac{12}{2} (16+20)
&= 216
∴ textual content{Space} &= 216 textual content{ mm}^2
finish{align*}

start{align*}
A &=2 pi occasions 4^2 + 2 pi occasions 4 occasions 9
&= 32 pi + 72 pi
&= 104 pi
∴ textual content{Floor space of can} &= 104 pi textual content{ cm}^2
finish{align*}

 (r = textual content{radius})

Substitute (r= frac{0.07}{2}) (because the radius is half of the diameter) into (A = 4 pi r^2):

start{align*}
A &= 4 pi occasions left( frac{0.07}{2} proper)^2
&= 4 pi occasions 0.035^2
&= 0.0049 pi
∴ textual content{Floor space of cell} &= 0.0049 pi textual content{ mm}^2
finish{align*}

To search out the world of the round base, substitute (r=5) into (A= pi r^2):

start{align*}
A &= pi occasions 5^2
&= 25 pi
∴ textual content{Base space} &= 25 pi textual content{ cm}^2
finish{align*}

Substitute (A= 25 pi) and (h=12) into (V= frac{1}{3}Ah):

start{align*}
V &=frac{1}{3} occasions 25 pi occasions 12
&= 100 pi
∴ textual content{Quantity of water wanted} &= 100 pi textual content{ cm}^3
&= 100 pi textual content{ mL}
finish{align*}

start{align*}
V &= frac{4}{3} pi left( frac{3}{2} proper) ^3
&= frac{9}{2} pi
∴ textual content{Quantity of mince wanted} &= frac{9}{2} pi textual content{ cm}^3
&= frac{9}{2} pi textual content{ g}
finish{align*}

n &= textual content{place of the time period}
a &= textual content{first time period: }T_1
d &= textual content{frequent distinction between phrases}
l &= textual content{final time period being added: }T_n
finish{align*}

start{align*}
T_{23} & = 3+(23-1) occasions 12
&= 3 + 22 occasions 12
&= 267
∴ textual content{23}^{rd} textual content{ time period: } & 267
finish{align*}

start{align*}
632 & =frac{16}{2} [2 times 2 +(16-1)d]
632 &= 8 occasions (4+ 15d)
632 &= 32+ 120d
600 &= 120d
∴ d &= 5
finish{align*}

start{align*}
T_1 & = a = 2 textual content{ (given)}
textual content{Since } T_n & = a+(n-1)d textual content{:}
T_2 &= 2 + 5
&= 7
T_3 &= 2+ 2 occasions 5
&= 12
T_4 &= 2 + 3 occasions 5
&= 17
textual content{∴ First 4 phrases of the sequence: }& 2 + 7 + 12 +17
finish{align*}

n &= textual content{variety of phrases being added}
a &= textual content{first time period: }T_1
r &= textual content{frequent ratio between phrases}
finish{align*}

Let’s take into account what this drawback would appear like visually.

The peak of every bounce conveniently types a geometrical sequence the place the frequent ratio r=0.75 and the primary time period a=x.

Therefore, (T_n=x occasions 0.75^{n-1})

Nevertheless, the gap that the ball travels shouldn’t be a easy geometric sequence. As you may see within the diagram above, aside from the primary bounce and last n bounce (the place the ball solely travels downwards or upwards), the ball travels twice the gap of the bounce peak because it travels up and down.

Therefore, we are able to discover the whole distance d travelled by the ball earlier than the nth bounce:

start{align*}
d &=  2 occasions [text{sum of the heights of the 1st, 2nd, 3rd, …, (n-2)th, (n-1)th and nth bounce}] -[text{height of the 1st bounce}] – [text{height of the nth bounce}]
&= 2 occasions (T_1+T_2+T_3+…+T_{n-2}+T_{n-1}+T_n) – T_1 – T_n
∴d &= 2 occasions S_n – T_1 – T_n
finish{align*}

On this query, we’re given the whole distance travelled by the ball earlier than the eighth bounce is 6 metres. So, substitute d=6 and  n=8 into ( d = 2 occasions S_n – T_1 – T_n ):

start{align*}
6 &= 2 occasions S_8 – T_1 – T_8 textual content{…………. (equation 1)}
finish{align*}

To search out S₈, substitute (a=x), (n=8) and (r=0.75) into (S_n=frac{a(1-r^n)}{1-r}):

start{align*}
S_8 &=frac{x(1-0.75^8)}{1-0.75}
&= 4[x(1-0.75^8)]
finish{align*}

To search out T₈, substitute (a=x) and (n=8) into (T_n=ar^{n-1}):

start{align*}
T_8 &= x occasions 0.75^{8-1}
&= x occasions 0.75^7
finish{align*}

Therefore, we are able to substitute (S_8=4[x(1-0.75^8)] ), (T_1= x ) and (T_8= x occasions 0.75^7 ) into equation 1: (6 = 2 occasions S_8 – T_1 – T_8):

start{align*}
6 &= 2 occasions 4[x(1-0.75^8)] – x – x occasions 0.75^7
6 &= 8[x(1-0.75^8)] – x – x occasions 0.75^7
6 &= x[8(1-0.75^8) – 1 – 0.75^7]
x &= frac{6}{8(1-0.75^8) – 1 – 0.75^7}
x &= 0.989 textual content{ (rounded to three decimal locations)}
∴ textual content{Top at which participant started bouncing ball} &= 0.989 textual content{ m}
finish{align*}

Partially a), we have been within the distance that the ball travels earlier than the eighth bounce. So, we used (n = 8). Partially b) although, we’re looking for how far the ball can journey if it was allowed to bounce without end. So right here, we let (n = ∞).

Substitute (n = ∞) into (d = 2 occasions S_n – T_1 – T_n):

start{align*}
d &= 2 occasions S_∞ – T_1 – T_∞
finish{align*}

Since:

start{align*}
T_1 &=a
&=0.989 textual content{ (from half a)}
S_∞ &= frac{a}{1-r},   textual content<1
&=  frac{0.989}{1-0.75}
& = 3.956
T_∞ &= 0 textual content{ (the ball doesn’t raise off the bottom on the ‘final’ bounce)}
finish{align*}

start{align*}
⇒ d &= 2 occasions S_∞ – T_1 – T_∞
&= 2 occasions 3.956 – 0.989 – 0
&= 6.92 textual content{ (rounded to 2 decimal locations)}
∴ textual content{Complete limiting distance} &= 6.92 textual content{ m}
finish{align*}