<u><em>Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.</em></u>
Answer:
Any value other than the values
will not be a solution of
.
Step-by-step explanation:
Considering the equation

Steps to solve the equation









As
![\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7Dx%5E3%3Df%5Cleft%28a%5Cright%29%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7Dx%3D%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%2C%5C%3A%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%5Cfrac%7B-1-%5Csqrt%7B3%7Di%7D%7B2%7D%2C%5C%3A%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%5Cfrac%7B-1%2B%5Csqrt%7B3%7Di%7D%7B2%7D)
![x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B8%7D%7D%2C%5C%3Ax%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Cfrac%7B-1%2B%5Csqrt%7B3%7Di%7D%7B2%7D%2C%5C%3Ax%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Cfrac%7B-1-%5Csqrt%7B3%7Di%7D%7B2%7D)
So,

Therefore,
Any value other than the values
will not be a solution of
.
Keywords: solution, value
Learn more about equation solution from brainly.com/question/1679491
#learnwithBrainly
It would be 3 because women’s meme
There is no one ordered pair that makes the equation true, this is a line which has infinitely many ordered pairs that are true...
2x+y=11 subtract 2x from both sides and you have the line in slope intercept form..
y=11-2x
For any real given value of x, you will get one real unique corresponding value for y, resulting in infinitely many unique (x,y) values, points, or ordered pairs, satisfying the given equation.
I assume that you were given "choices" to pick from because of the above reality. So if any of your choices does not satisfy y=11-2x it is not an ordered pair satisfying the equation.
Answer:
r < -12 and r > -6
Step-by-step explanation:
x^2 in the function that you typed is x * x ( x multiplied by x). It represents a second degree because there are only 2 "x"s multiplied together. If there were 3 "x"s multiplied together, we would have x^3. Since the highest exponent in this function is 2 (in x^2), we call this function a 2nd degree polynomial.