59) Pls answer asap BEST ANSWER BRAINLIEST ridiculous answers will be deleted

Q(–1, a) is 4 5 units away from P(3, –2). Find all possible values of a.

(there are two values)

The possible values of a is 43

Step-by-step explanation:

Using the formula for calculating the distance between 2 points.

D = √(x2-x1)²+(y2-y1)²

Given

D = 45 units

The coordinates Q(–1, a) and P(3,2)

45 = √(3-(-1))²+(-2-a)²

45 = √4²+(-2-a)²

45 = √16+(-2-a)²

45² = 16+(-2-a)²

45² = 16+4+4a+a²

45² = 20+4a+a²

a²-4a+20-2025 = 0

a²+4a-2005 = 0

a = -4±√16+4(2005)/2

a = -4±√16+8020/2

a = -4±√8036/2

a =  -4±90/2

a = -4+90/2

a = 86/2

a = 43  

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