K+2 ≥ 5 i need the answer pls

Mathematics

2 answers:

nadezda [96]3 years ago

8 0

Answer: dont know

Step-by-step explanation:

Semmy [17]3 years ago

6 0

K=3 or higher, I would need to know the options to tell you an exact answer

You might be interested in

(2x+3)²=5.find the roots of the equation

bija089 [108]

Answer:

-2.61803398875, -0.38196601125

(-3±√5)/2

Step-by-step explanation:

First, expand. (2x+3)² = 4x²+12x+9=5

4x²+12x+4 = 0

Quadratic equations take the form ax²+bx+c = 0

In this case,

a = 4

b = 12

c = 4

You can solve with the equation x=(-b±√b²-4ac)/2a

Plug in:

(-12±√144-64)/8

(-12±√80)/8

(-12±√2^4×5)/8

(-12±4√5)/8

(-3±√5)/2

Final answer x = (-3±√5)/2

If you want decimal form: -2.61803398875, -0.38196601125

4 0

2 years ago

Which are the solutions of x^2 = –11x + 4?

strojnjashka [21]

This is a quadratic equation, i.e. an equation involving a polynomial of degree 2. To solve them, you must rearrange them first, so that all terms are on the same side, so we get

$x^2 + 11x - 4 = 0$

i.e. now we're looking for the roots of the polynomial. To find them, we can use the following formula:

$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2}$

where $x_{1,2}$ is a compact way to indicate both solutions $x_1$ and $x_2$ , while $a,b,c$ are the coefficients of the quadratic equation, i.e. we consider the polynomial $ax^2+bx+c$ .

So, in your case, we have $a=1,\ \ b=11,\ \ c=-4$

Plug those values into the formula to get

$x_{1,2} = \frac{-11\pm\sqrt{121+16}}{2} = \frac{-11\pm\sqrt{137}}{2}$

So, the two solutions are

$x_1 = \frac{-11+\sqrt{137}}{2}$

$x_2 = \frac{-11-\sqrt{137}}{2}$