Can a triangle be formed with side lengths: 10, 12, 22

1 answer:

8 0

No. A squared plus b squared must equal c squared (Pythagorean theorem). 10 squared is 100, 12 squared is 144, they add up to 244 but 22 squared is 484…..much too large

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What is the area of a parallelogram whose vertices are A(-12,2) B(6,2) C(-2,-3, D(-20,-3)

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3 years ago

Anyone know how to prove these are equal to one another? I also need all the steps.

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Answer:

Step-by-step explanation:

it can be simplified to $\frac{sec \theta}{cot\theta+tan\theta} =sin\theta$ . if you multiply each side by the denominator, it becomes $sec\theta=\frac{sin^2\theta}{cos\theta} + cos\theta$ since $tan\theta=\frac{sin\theta}{cos\theta}$ and cotan is the reciprocal of that. Because $sec\theta=\frac{1}{cos\theta}$ so multiplying each side gives $1=sin^2\theta+cos^2\theta$ , and $sin^2\theta+cos^2\theta$ is equal to 1.