Answer:
x=11, y=2
Step-by-step explanation:
We can set 1 equal to x-5y and then solve for x. and y.
x = 5y+1
y = x-1/5
We can use this information and plug back in the values for 3y-x or x-5y.
We can set -3 = 3y-x or 1 = x-5y.
To solve for x using -3 = 3y-x we can swap the values of x and y which would make it -3 = 3(x-1/5)-5(x-1/5)+1.
We can do a bit of algebra which would get us x = 11.
Knowing that y = x-1/5 we can plug in 11 for x. y = 11-1/5.
y=2
x=11, y=2
Answer:
w=4
Step-by-step explanation:
The vertex (minimum) of the quadratic ax² +bx +c is located at x=-b/(2a). This means the minimum value of f(x) will be found at x = -3/(2*1) = -1.5.
Since the vertex of the quadratic is less than 0, the maximum value of the quadratic will be found at x=2, the end of the interval farthest from the vertex.
On the given interval, ...
the absolute minimum value of f is f(-1.5) = ln(1.75) ≈ 0.559616
the absolute maximum value of f is f(2) = ln(14) ≈ 2.639057
7.86 x 4 is 36.156 because you have to line up the decimals and multiply them and then you get 36.156