A particle moves along the curve LaTeX: y=\sqrt{1+x^3}y = 1 + x 3. As it reaches the point (2 , 3), the y-co-ordinate is increasing at a rate of 4 cm/s. How fast is the x-co-ordinate of the point changing at that instant? Round your answer to 2 decimal places.

2 cm/s

Step-by-step explanation:

y=√{1+x³}

x and y are dependent on t and so;

y(t) =√{1+x(t)³}

Thus;

dy/dx = y'(t) = [3x(t)²•x'(t)]/(2√{1+x(t)³})

We know that;

y=√{1+x³}

Thus;

y'(t) = [3x(t)²•x'(t)]/2y

We are given y' (t) at this instant t(o), thus; y '(t(o)) = 4

and also at that very same t(o);

x(t(o)) = 2 and y(t(o)) = 3

Thus;

Using, y'(t) = [3x(t)²•x'(t)]/2y, we obtain;

4 = [3 x 2² • x'(t)]/2 x 3

4 x 2 x 3 = 12(x'(t))

24 = 12 (x'(t))

(x'(t)) = 24/12 = 2 cm/s

correct answer:

b.(19,-2)

answer: 33 ( approx)

step-by-step explanation:

since, in the linear regression formula, the slope is the a in the equation y = b + ax ( by comparing it with the general equation of line y = mx + c )

where,

here given table is,

price ($), x   :           18         23       25       29       30       33

pairs sold, y :         16         13         10       8         6       2

so, ,   ,   ,

and   ,

thus, slop of the given line is,

⇒  

⇒   ≈ 33