Vector curve problem. Part of a cross-country skier's path can be described with the vector function r = <2 + 6t 2 + cos(t), (15 − t)(1 − sin(t))> for 0 ≤ t ≤ 15 minutes, with x and y measured in meters. The derivatives of these functions are given by x′(t) = 6 − 2sin(t) and y′(t) = −15cos(t) + tcos(t) − 1 + sin(t).

Find the acceleration vector of the skier when the skier's horizontal position is x = 60.

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation such as light—can escape from it.[1] The theory of general relativity predicts that a sufficiently compact mass can deform spacetime to form a black hole.[2][3]

The boundary of the region from which no escape is possible is called the event horizon. Although the event horizon has an enormous effect on the fate and circumstances of an object crossing it, according to general relativity it has no locally detectable features.[4] In many ways, a black hole acts like an ideal black body, as it reflects no light.[5][6] Moreover, quantum field theory in curved spacetime predicts that event horizons emit Hawking radiation, with the same spectrum as a black body of a temperature inversely proportional to its mass. This temperature is on the order of billionths of a kelvin for black holes of stellar mass, making it essentially impossible to observe directly.

Objects whose gravitational fields are too strong for light to escape were first considered in the 18th century by John Michell and Pierre-Simon Laplace.[7] The first modern solution of general relativity that would characterize a black hole was found by Karl Schwarzschild in 1916, although its interpretation as a region of space from which nothing can escape was first published by David Finkelstein in 1958. Black holes were long considered a mathematical curiosity; it was not until the 1960s that theoreti

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step-by-step explanation:

the system of inequalities is

step-by-step explanation:

step 1

find the equation of the dashed line

we have the points

(0,0) and (3,1)

the line passes through the origin, so, is a direct variation

find the slope

the equation of the dashed line is

the solution of the inequality a s the shaded area above the dashed line

therefore

the equation of the inequality a is

step 2

find the equation of the solid line

we have the points

(4,0) and (0,4)

find the slope

the equation of the solid line is

the solution of the inequality b s the shaded area below the solid line

therefore

the equation of the inequality b is

step 3

the system of inequalities is