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3 years ago

What is the general form of the equation of the given circle with center A

1 answer:

7 0

If center A is at the point (h, k), the formula for the circle is r^2 = (x-h)^2+(y-k)^2, where r is the radius of the circle

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19 . 2.667 times larger

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2 years ago

Say that you leave montevideo at 6:20 p.m. on sunday to fly to jerusalem. when you arrive in jerusalem, the local time is 10:00

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Its A. Five hours ahead :)

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2 years ago

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What is 2+2. Im in highschool and this is hard.

Murrr4er [49]

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2 years ago

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Please help look at attached image below need an answer asap

NNADVOKAT [17]

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8 0

2 years ago

Simplify the expression using trigonometric identities:

Anastaziya [24]

The simplified expression is the one in option B.

<h3>How to simplify the expression?</h3>

Remember that:

$sec(x) = 1/cos(x)\\\\cos(x) = cos(-x)$

With these two, we can rewrite the numerator as:

$sec(x)*sin(-x) + tan(-x)\\\\\frac{1}{cos(x)}*sin(-x) + tan(x)\\\\\frac{1}{cos(-x)}*sin(-x) + tan(-x)\\\\2*tan(-x)$

Replacing that in the given expression, we have:

$\frac{2*tan(-x)}{1 + sec(-x)} = \frac{2*tan(-x)}{1 + 1/cos(-x)}$

Now, if we multiply and divide by cos(-x), we get:

$\frac{2*tan(-x)}{1 + 1/cos(-x)} = \frac{2*tan(-x)}{1 + 1/cos(-x)}*\frac{cos(-x)}{cos(-x)} \\\\= \frac{2*sin(-x)}{1 + cos(-x)}$

Now remember that:

$cos(x) = cos(-x)\\sin(-x) = -sin(x)$

With these two properties:

$\frac{2*sin(-x)}{1 + cos(-x)} = \frac{-2*sin(x)}{1 + cos(x)}$

Then the correct option is B.

If you want to learn more about trigonometric identities:

brainly.com/question/7331447

#SPJ1

3 0

1 year ago