A lake near the Arctic Circle is covered by a 2-meter-thick sheet of ice during the cold winter months. When spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a constant rate. After 3 weeks, the sheet is only 1.25 meters thick. Let y represent the ice sheet's thickness ( in meters) after x weeks. Complete the equation for the relationship between the thickness and number of weeks.
Answer:
y = 2 - 0.25x
Step-by-step explanation:
From the question, the initial thickness of the ice sheet = 2 meters,
After 3 weeks, the thickness of ice sheet reduced to 1.25 meters
Hence, the difference in the thickness in 3 weeks is calculated as:
2m - 1.25m = 0.75m
The amount of changes that occurred in 3 weeks is given as:
= 0.75/3 = 0.25 meters,
We are told that, the ice is melting with the constant rate.
Therefore, the equation for the relationship between the thickness and number of weeks is given as:
y = 2 - 0.25x
Answer:
<h2> no, because the remainder is 126</h2>
Step-by-step explanation:
if x+3 is a factor, then -3 is a root of expression, and the remainder would be 0
calculating remainder:

Answer:
193900
Step-by-step explanation:
thx for points
Answer:
mode
Step-by-step explanation:
mode is the number that has the highest occurrences in a set of numbers
You do the implcit differentation, then solve for y' and check where this is defined.
In your case: Differentiate implicitly: 2xy + x²y' - y² - x*2yy' = 0
Solve for y': y'(x²-2xy) +2xy - y² = 0
y' = (2xy-y²) / (x²-2xy)
Check where defined: y' is not defined if the denominator becomes zero, i.e.
x² - 2xy = 0 x(x - 2y) = 0
This has formal solutions x=0 and y=x/2. Now we check whether these values are possible for the initially given definition of y:
0^2*y - 0*y^2 =? 4 0 =? 4
This is impossible, hence the function is not defined for 0, and we can disregard this.
x^2*(x/2) - x(x/2)^2 =? 4 x^3/2 - x^3/4 = 4 x^3/4 = 4 x^3=16 x^3 = 16 x = cubicroot(16)
This is a possible value for y, so we have a point where y is defined, but not y'.
The solution to all of it is hence D - { cubicroot(16) }, where D is the domain of y (which nobody has asked for in this example :-).
(Actually, the check whether 0 is in D is superfluous: If you write as solution D - { 0, cubicroot(16) }, this is also correct - only it so happens that 0 is not in D, so the set difference cannot take it out of there ...).
If someone asks for that D, you have to solve the definition for y and find that domain - I don't know of any [general] way to find the domain without solving for the explicit function).