X² - 15 x + 36 = 0
x² - 12x - 3x + 36 = 0
x( x - 12) - 3( x - 12) = 0
(x - 3)( x - 12) = 0

x - 3 = 0
x = 3
or
x - 12 = 0
x = 12

The solutions are 3 and 12

7 0

3 years ago

Please help me out with this please

ki77a [65]

Answer:

P = 9 is the max value

Step-by-step explanation:

Sketch

2x + 4y = 10

with x- intercept = (5, 0) and y- intercept (0, 2.5)

x + 9y = 12

with x- intercept = (12, 0) and y- intercept = (0, $\frac{4}{3}$ )

Solve

2x + 4y = 10 and x + 9y = 12 to find the point of intersection at (3, 1)

The region corresponding to the solution of the system of constraints

Has vertices at (0, $\frac{4}{3}$ ), (0, 0) , (5, 0) and (3, 1)

Now evaluate the objective function at each vertex.

(0, 0) can be excluded as it will not give a maximum

(5, 0) → P = 5 + 0 = 5

(0, $\frac{4}{3}$ ) → 0 + 8 = 8

(3, 1) → 3 + 6(1) = 3 + 6 = 9 ← maximum value

Thus the maximum value is 9 when x = 3 and y = 1

7 0

3 years ago

What is the phase “hi friend” in binary code

mihalych1998 [28]

01101000 01101001 00100000 01100110 01110010 01101001 01100101 01101110 01100100

6 0

3 years ago

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