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natta225 [31]
4 years ago
5

Prove that a cyclic parallelogram is a rectangle.

Mathematics
1 answer:
ZanzabumX [31]4 years ago
4 0

Answer:

Step-by-step explanation:

let ABCD be cyclic parallelogram.

Then ∠A=∠C

as it is cyclic so ∠A+∠C=180

∠C+∠C=180

2∠ C=180

∠C=90

So ∠A=90

similarly ∠B=∠D=90

so it is a rectangle.

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Solve the following equation for 0° ≤ θ < 360°. Use the "^" key on the keyboard to indicate an exponent. For example, sin2x w
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<u>Answer-</u>

\boxed{\boxed{x=90^{\circ},270^{\circ}}}

<u>Solution-</u>

The given equation-

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As

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Putting this,

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\Rightarrow 3\sin^2 x-3=0

\Rightarrow 3\sin^2 x=3

\Rightarrow \sin^2 x=1

\Rightarrow \sin x=\sqrt1

\Rightarrow \sin x=\pm 1

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\Rightarrow x=\sin^{-1}(1),\ x=\sin^{-1}(-1)

\Rightarrow x=90^{\circ}+n360^{\circ},\ x=270^{\circ}+n360^{\circ}

Where n=0, 1, 2, ......

As given 0° ≤ x < 360°, so putting n = 0

\Rightarrow x=90^{\circ},\ x=270^{\circ}


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4 years ago
Which equation in point-slope form contains the point (2, 3) and has slope 2?
sergiy2304 [10]
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Option A
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Check if slope = 2
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Answer:  Yes, Slope is 2

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When x = 2
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Answer: No, it does not passed through (2, 3)

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Option B
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y - 3 = 2(x - 2)
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Check if it passed through (2, 3)
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when x = 2
y = 2(2) - 1
y = 4 - 1
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Answer : Yes, it passed through (2, 3)

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Answer: (B) <span>y – 3 = 2(x – 2)
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