Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

<h3>Given information-</h3>

The given function is,

$h(x)=x^2+1$

$k(x)=x-2$

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

$(h+k)(x)=h(x)+k(x)$

$(h+k)(x)=x^2+1+x-2$

$(h+k)(2)=2^2+1+2-2$

$(h+k)(x)=5$

2)The value of the function (h-k)(3),

$(h-k)(x)=h(x)-k(x)$

$(h-k)(x)=x^2+1-x+2$

$(h-k)(3)=3^2+1-3+2$

$(h-k)(x)=9$

3) The value of the function 3h(2)+2k(3)

$3h(x)+2k(x)=3x^2+3+2x-2\times 2$

$3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2$

$3h(2)+2k(3)=17$

Hence the value of the given functions are,

$(h+k)(x)=5$

$(h-k)(x)=9$

$3h(2)+2k(3)=17$

Learn more about the quadratic equation here;

brainly.com/question/2263981

4 0

2 years ago

How would the following triangle be classified?

Anna [14]

Answer:

hope this helpful

Step-by-step explanation:

isosceles right

because A light ray incident at 90° at the first face emerges at the same angle. The diagram shows five isosceles right-angled prisms. A light ray incident at 90° at the first face emerges at the same angle with the normal from the last face.

8 0

2 years ago