Determine the quadrant(s) in which (x, y) could be located. (Select all that apply.) x <0 and y < 0
Quadrant 1
Quadrant II
Quadrant III
Quadrant IV
none of these

D. Quadrant IV

Step-by-step explanation: Numbers with a positive or negative sign are referred to be directed numbers. These numerals are easily found on the number line. 2, 5, -1, -3, -7, 9, and so on are some examples.

As a result: x > 0 and y 0 are both true.

Consider the case of specific directed numbers for x and y. Assume x = 5 and y = -2, and the following is true: (5, -2). In a Cartesian plane, this point can be found in the fourth quadrant.

As a result, in a Cartesian plane, the quadrant in which (x, y) might be found is Quadrant IV.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!