Evaluate ƒ(x) = 3|x – 2| + 1 for ƒ(–2) and ƒ(1).

ƒ(x) = 3|x – 2| + 1

To find f(-2) substitute - 2 into f(x)

That's

f(-2) = 3| - 2 - 2 | + 1

= 3| - 4| + 1

But absolute value of any number is positive including negative numbers

That's

| - 4 | = 4

So we have

3(4) + 1

12 + 1

To find f(1) substitute 1 into f(x)

That's

f(1) = 3 | 1 - 2| + 1

= 3 | - 1| + 1

But | - 1| = 1

= 3(1) + 1

= 3 + 1

Hope this helps you

f(-2) =13

f(-1) = 4

Step-by-step explanation:

ƒ(x) = 3|x – 2| + 1

Let x = -2

ƒ(-2) = 3|-2 – 2| + 1

       = 3 | -4| +1

Taking the absolute value

    = 3*4 +1

    = 12 +1 = 13

Let x = 1

ƒ(1) = 3|1 – 2| + 1

       = 3 | -1| +1

Taking the absolute value

    = 3*1 +1

    = 3 +1 = 4

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations: