2 years ago

Find the range of values for x See pic

Mathematics

1 answer:

Dvinal [7]2 years ago

8 0

Answer:

Step-by-step explanation:

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2. Solve the equation using the quadratic formula. -4x2 + 40x + 16 = 0 Page 1 of 2

kumpel [21]

The solution to given quadratic equation is x = -0.3851 or x = 10.3851

Solution:

Given equation is:

$-4x^2 + 40x + 16 = 0$

To find: solve by quadratic equation formula

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text{ For} -4x^2 + 40x + 16 = 0 \text{ we have } a = -4 ; b = 40 ; c = 16$

Substituting the values of a = -4 ; b = 40 ; c = 16 in above quadratic formula we get,

$\begin{aligned}&x=\frac{-40 \pm \sqrt{40^{2}-4(-4)(16)}}{2 \times-4}\\\\&x=\frac{-40 \pm \sqrt{1600+256}}{-8}\\\\&x=\frac{-40 \pm 43.081}{-8}\\\\&x=\frac{-40+43.081}{-8} \text { or } x=\frac{-40-43.081}{-8}\\\\&x=-0.3851 \text { or } x=10.3851\end{aligned}$

Thus the solution to given quadratic equation is x = - 0.3851 or x = 10.3851

8 0

2 years ago

Is the point (1,-3) a solution to the system? show all your work y>x-1 y >-x+2

vagabundo [1.1K]

Answer:

To determine if the point is a solution, substitute the x value into the right side of the inequality and compare the result with the y value of the point.

- x + 2 = - 1 + 2 = 1

Now - 3 < 1

Thus (1, - 3) is not a solution

Step-by-step explanation:

6 0

2 years ago