The value of 2 would be 0.02, or 0.020, or two hundredths.
Answer:
Impossible
Step-by-step explanation:
In 5x^2-4x+3=0, standard form, substitute these values in the quadratic formula:
a = 5; b = -4; c = 3
The quadratic formula is ![x = \frac{ -b ± \sqrt{b^{2} - 4ac}}{2a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B%20-b%20%C2%B1%20%5Csqrt%7Bb%5E%7B2%7D%20-%204ac%7D%7D%7B2a%7D)
(ignore the weird capital A)
Substitute a b and c:
![x = \frac{-(-4) ± \sqrt{(-4)^{2} - 4(5)(3)}}{2(5)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-%28-4%29%20%C2%B1%20%5Csqrt%7B%28-4%29%5E%7B2%7D%20-%204%285%29%283%29%7D%7D%7B2%285%29%7D)
Simplify:
![x = \frac{4) ± \sqrt{16 - 60}}{10}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B4%29%20%C2%B1%20%5Csqrt%7B16%20-%2060%7D%7D%7B10%7D)
Because
is the square root of a negative number, the answer would be imaginary.
Therefore, there are not solutions to this equation.
A solution is the same as the roots or zeroes, where the graph would cross the x-axis when graphed.
The graph never meets the x-axis. It looks like this:
<em>Answer</em>
<em>To represent -5/3 on number line. Solution: </em>
<em>Take </em>a line X'X and a suitable point zero. On the right of 0 every 9t<em>h -s you indicate 1,2,3 etc. On the left of 0, every 9th -s you indicate -1 Or1', -2 or 2' , -3 or 3'.</em>
Answer: girllll ion know
Step-by-step explanation:
It is a because the model is up then b so it will be a