The temperature at a point (x, y) is t(x, y), measured in degrees celsius. a bug crawls so that its position after t seconds is given by the following function, where x and y are measured in centimeters. x = sqrt(2 + t), y = 5 + 1/14 t the temperature function satisfies tx(4, 6) = 8 and ty(4, 6) = 4. how fast is the temperature rising on the bug's path after 14 seconds?

The rate of change of temperature is 1.29 degree Celsius per second.  

Step-by-step explanation:

We are given the following information in the question:

The temperature at a point (x, y) is T(x, y), measured in degrees Celsius where x and y are measured in centimeters.

We have to find the rate at which the temperature is rising on the bug's path after 14 seconds.

At t = 14 seconds, we have,

To find rate of change of temperature, we differentiate,

Thus, the rate of change of temperature is 1.29 degree Celsius per second.

26 is 33.33% of 78

step-by-step explanation:

26 is what percent of 78?

26 is p% of 78

equation: y = p% * x

solving our equation for p

p% = y/x

p% = 26/78

p = 0.3333

convert decimal to percent:

p% = 0.3333 * 100 = 33.33%