Question

math
questions:

1. if x = -2 and y = 3 evaluate ——> 6x³ + 3x² - 2y
2. select the constant term —> 15x + 2 —> 2
3. a) evaluate (5)³

b) (-12)⁰

1. 3(2x + 5) = -4 (x + 10)

solve —>

1. \\(\frac{-3x}{5}\\) = 12

Explanation:

Step1: Substitute x=-2, y=3

$6(-2)^3 + 3(-2)^2 - 2(3)$

Step2: Calculate each term

$6(-8) + 3(4) - 6 = -48 + 12 - 6$

Step3: Simplify the expression

$-48 + 12 - 6 = -42$

Step1: Identify constant term

Constant term: 2

Step1: Evaluate positive cube

$5^3 = 5\times5\times5 = 125$

Step2: Evaluate zero exponent

$(-12)^0 = 1$ (Any non-zero number to 0 power is 1)

Step1: Expand both sides

$6x + 15 = -4x - 40$

Step2: Collect x terms left

$6x + 4x = -40 - 15$

Step3: Simplify and solve for x

$10x = -55$
$x = \frac{-55}{10} = -\frac{11}{2}$

Answer:

1. $-42$
2. $2$
3. a) $125$; b) $1$
4. $x = -\frac{11}{2}$
5. $x = -20$