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Cloud [144]
2 years ago
11

I really don't know how to do this-

Mathematics
2 answers:
shepuryov [24]2 years ago
6 0

The sum of the interior angles of any quadrilateral is 360°.

So,

90° + 90° + 52° + x = 360°

=> 232° + x = 360°

=> x = 360° - 232°

=> x = 128°

Elodia [21]2 years ago
6 0

Answer:

x = 128

Step-by-step explanation:

All angles on a trapezoid must add up to 360 degrees. So we have 3 of our 4 angles so we can solve this as an equation

90 + 90 + 52 + x = 360

232 + x = 360

x = 128

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7 0
3 years ago
Please help me it is a quiz
fredd [130]

Answer:

Step-by-step explanation: I CAN HELP YOU what is the question

4 0
2 years ago
 If the trapezoid below is reflected across the x-axis, what are the coordinates of B'?
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9 0
3 years ago
Read 2 more answers
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4 0
3 years ago
Prove the trigonometric identity
Annette [7]

Answer:

Proved See below

Step-by-step explanation:

Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started

For this question we need to know the following identities:

1+tan^{2}x=sec^2x\\\\1+cot^2x=cosec^2x\\\\sin^2x+cos^2x=1

Lets solve the bottom most part first:

1-\frac{1}{1-sec^2x} \\\\

Take LCM

1-\frac{1}{1-sec^2x} \\\\\frac{1-sec^2x-1}{1-sec^2x} \\\\\frac{-sec^2x}{1-sec^2x} \\\\\frac{-(1+tan^2x)}{-tan^2x}

now break the LCM

\frac{-1}{-tan^2x}+\frac{-tan^2x}{-tan^2x}\\\\\frac{1}{tan^2x}+1\\\\cot^2x+1

because 1/tan = cot x

and furthermore,

cot^2x+1\\cosec^2x

now we solve the above part and replace the bottom most part that we solved with cosec^2x

\frac{1}{1-\frac{1}{cosec^2x} } \\\\\frac{1}{1-sin^2x} \\\\\frac{1}{cos^2x}\\\\sec^2x

Hence proved! :D

4 0
3 years ago
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