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3 years ago

Given: Circle A externally tangent to Circle B.

Mathematics

1 answer:

bulgar [2K]3 years ago

4 0

Answer:

Line r

Explanation:

A common internal tangent is a tangent of both circles in which the intersection of the tangent and the line segment joining the centers is not empty.

This means that the tangent and the line segment joining the centers must intersect.

Line s does not intersect AB, the line segment joining the circles. This means line s is not the common internal tangent.

Segment AB is the segment joining the centers; it is not tangent to the circles.

Line r is tangent to both circles and intersects AB; this is the common internal tangent.

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Read 2 more answers

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Mandarinka [93]

Answer:

The standard form of the equation is 49t² - 75t + 3 = 0

The solution of the equations are 1.49 and 0.041

Step-by-step explanation:

* Lets explain how to solve the problem

- The standard form of the quadratic equation is ax² + bx + c = 0,

where a , b , c are constant and a can not be 0

∵ The quadratic equation is -4.9t² + 7.5t + 1.8 = 2.1

- Lets make the left hand side equal to 0

∵ -4.9t² + 7.5t + 1.8 = 2.1 ⇒ subtract 2.1 from both sides

∴ -4.9t² + 7.5t - 0.3 = 0 ⇒ multiply each term by -10

∴ 49t² - 75t + 3 = 0

* The standard form of the equation is 49t² - 75t + 3 = 0

∵ ax² + bx + c = 0

∴ a = 49 , b = -75 , c = 3

- Lets use the formula $x=\frac{-b+-\sqrt{b^{2}-4ac}}{2a}$ to solve

the equation

∴ $x=\frac{-(-75)+\sqrt{(-75)^{2}-4(49)(3)}}{2(49)}=1.49$

∴ $x=\frac{-(-75)-\sqrt{(-75)^{2}-4(49)(3)}}{2(49)}=0.041$

* The solution of the equations are 1.49 and 0.041

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