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Paraphin [41]
4 years ago
13

Help pls this is a missing assignment

Mathematics
2 answers:
murzikaleks [220]4 years ago
5 0

Always. The answer is always

Paul [167]4 years ago
4 0
<h2><u>Ans</u></h2><h2><u>your ANSWER IS NONE </u></h2>

<h2><u><em>PLZZZZ MARK ME AS BRAINLIST </em></u></h2>
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Tomtit [17]

Answers:

A ' = (-2, -3)

B ' = (0, -3)

C ' = (-1, 1)

=======================================================

Explanation:

To apply an x axis reflection, we simply change the sign of the y coordinate from positive to negative, or vice versa. The x coordinate stays as is.

Algebraically, the reflection rule used can be written as (x,y) \to (x,-y)

Applying this rule to the three given points will mean....

  • Point A = (-2, 3) becomes A ' = (-2, -3)
  • Point B = (0, 3) becomes B ' = (0, -3)
  • Point C = (-1, -1) becomes C ' = (-1, 1)

The diagram is provided below.

Side note: Any points on the x axis will stay where they are. That isn't the case here, but its for any future problem where it may come up. This only applies to x axis reflections.

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4 years ago
म़ीरा ने किस को अपऩी जाग़ीर बताया है?class 10​
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कृष्ण‌ को marks as brain list

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3 years ago
____ is a line that passes through exactly one point on a circle
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3 years ago
Csc^3x-csc^2x-cscx+1=cot^2x(cscx-1)​
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Verify the identity:

csc^3x-csc^2x-cscx+1\qquad =cot^2x(cscx-1)\\\\csc^2x(cscx-1)-1(cscx-1) =\quad \downarrow\\\\(csc^2x-1)(cscx-1)\quad \qquad \ =\quad \downarrow\\\\\bigg(\dfrac{1}{sin^2x}-\dfrac{sin^2x}{sin^2x}\bigg)(cscx-1)\ =\quad \downarrow\\\\\\\bigg(\dfrac{1-sin^2x}{sin^2x}\bigg)(cscx-1)\qquad \ =\quad \downarrow\\\\\\\bigg(\dfrac{cos^2x}{sin^2x}\bigg)(cscx-1)\qquad \qquad =\quad \downarrow\\\\\\cot^2x(cscx-1)\qquad \qquad \qquad =cot^2x(cscx-1)\qquad \checkmark

5 0
3 years ago
Read 2 more answers
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