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neonofarm [45]
3 years ago
6

A sequence of translations maps ∆ABC to ∆A’B’C’.

Mathematics
1 answer:
Len [333]3 years ago
3 0

Answer:

(a) Co-ordinate rule is x'=x+6 and y'=y-8

(b) Co-ordinates of B' and C' are (3,-8) and (5,-5) respectively.

Step-by-step explanation:

(a)

Here, the co-ordinates of A(-3,4) are translated to A'(3,-4).

For the co-ordinates A and A',  x'-x=3-(-3)=3+3=6 and y'-y=-4-4=-8

So, x value of A has shifted to right by 6 units and y value of A has shifted 8 units down.

Hence, the co-ordinate rule that maps ΔABC onto ΔA'B'C' is:

x'=x+6 and y'=y-8.

(b)

Using the co-ordinate rule, we can find the co-ordinates of B' and C'.

For B, x=-3 and y=0.

So, x' of B' is x'=x+6=-3+6=3

And, y' of B' is y'=y-8=0-8=-8.

Therefore, co-ordinates of B' are (3,-8).

For C, x=-1 and y=3.

So, x' of C' is x'=x+6=-1+6=5

And, y' of C' is y'=y-8=3-8=-5.

Therefore, co-ordinates of C' are (5,-5).

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2 years ago
Find (a) arc length and (b) Area of a sector.
barxatty [35]

Answer:

a) 29.45 cm (2 dp)

b) 220.89 cm² (2 dp)

Step-by-step explanation:

<u>Formula</u>

\textsf{Arc length}=r \theta

\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta

\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}

<u>Calculation</u>

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\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 15\left(\dfrac{5 \pi}{8}\right)\\& = \dfrac{75}{8} \pi \\& = 29.45\: \sf cm\:(2\:dp)\end{aligned}

\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(15^2) \left(\dfrac{5 \pi}{8}\right)\\\\& = \dfrac{225}{2}\left(\dfrac{5 \pi}{8}\right)\\\\ & = \dfrac{1125}{16} \pi \\\\& = 220.89 \: \sf cm^2\:(2\:dp)\end{aligned}

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   Graham      =        Max

-50x + 14,040 = 20x + 12,500

<u>+50x               </u>   <u>+50x               </u>

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<u>          -12,500  </u>  <u>         -12.500 </u>

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