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zalisa [80]
3 years ago
7

Pls help to solve this ​

Mathematics
2 answers:
IRISSAK [1]3 years ago
5 0

Answer:

\huge\mathfrak\pink{S} \huge\mathfrak\purple{o} \huge\mathfrak\blue{l} \huge\mathfrak\red{u} \huge\mathfrak\green{t} \huge\mathfrak\pink{i}\huge\mathfrak\purple{o}\huge\mathfrak\red{n}\huge\mathfrak\orange{:}\

now

we have

P(x,y)⇢reflection under x axis⇢P'(x,-y)

P(8,10)⇢reflection under x axis⇢P'(8,-10)

for point (8,10) image is:(8,-10).

3241004551 [841]3 years ago
3 0

Answer:

\underline{ \underline{ \large{ \pink{ \tt{E \: X \: P \: L \: A \: N \: A \: T \: I \: O \: N}}} }}:

✧ We know , Reflection of a point on x - axis remains the x - co-ordinate same but the sign of y - coordinate changes i.e P ( x , y ) ⇢ P' ( x , - y ). So, the image of the point ( 8 , 10 ) under the reflection on x - axis is :

A ( 8 , 10 ) ⇢A' ( 8 , -10 )

☄ Hope I helped! ♡

☃ Have a wonderful day/ night ! ♨

\underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}} ✎

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

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  • right: no

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I need help figuring out what I did wrong
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Answer:

D

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Note that \frac{5\pi }{12} = \frac{\pi }{4} + \frac{\pi }{6} , thus

sin(\frac{5\pi }{12})

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3 years ago
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\qquad\qquad\huge\underline{{\sf Answer}}

let's evaluate ~

\qquad \sf  \dashrightarrow \: - 6 - 5(6x - 4)

\qquad \sf  \dashrightarrow \: - 6 - [(5 \cdot6x) - (5 \cdot4)]

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I hope you understood the whole procedure ~

7 0
2 years ago
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8 0
4 years ago
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gtnhenbr [62]

Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

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Step-by-step explanation:

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