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lianna [129]
3 years ago
15

If x^2= 20 what is the value of x

Mathematics
2 answers:
guajiro [1.7K]3 years ago
3 0

x^2 = 20.

sqrt(x^2) = sqrt(20)

x = +- 2sqrt(5)

Answer:

2\sqrt{5} and -2\sqrt{5}

OleMash [197]3 years ago
3 0

Answer:

4.472

Step-by-step explanation:

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5 0
4 years ago
Part C
Kay [80]

From the inscribed circle, AE and AD are the radius of the circle while BE and BD are the tangents.

<h3>What is an inscribed circle?</h3>

An inscribed circle simply means the largest circle that's contained in a triangle.

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Learn more about inscribed circle on:

brainly.com/question/12199537

8 0
2 years ago
√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1
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Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}


\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s

7 0
3 years ago
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