Answer:
![\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathrm%7BRange%5C%3Aof%5C%3A%7Dx%5E2%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Af%5Cleft%28x%5Cright%29%5Cge%20%5C%3A0%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%5B0%2C%5C%3A%5Cinfty%20%5C%3A%29%5Cend%7Bbmatrix%7D)
The graph is also attached below.
Step-by-step explanation:
Given the function
![y=x^2](https://tex.z-dn.net/?f=y%3Dx%5E2)
- We know that the range of a function is the set of values of the dependent variable for which a function is defined.
![\mathrm{For\:a\:parabola}\:ax^2+bx+c\:\mathrm{with\:Vertex}\:\left(x_v,\:y_v\right)](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3Aa%5C%3Aparabola%7D%5C%3Aax%5E2%2Bbx%2Bc%5C%3A%5Cmathrm%7Bwith%5C%3AVertex%7D%5C%3A%5Cleft%28x_v%2C%5C%3Ay_v%5Cright%29)
![\mathrm{If}\:a](https://tex.z-dn.net/?f=%5Cmathrm%7BIf%7D%5C%3Aa%3C0%5C%3A%5Cmathrm%7Bthe%5C%3Arange%5C%3Ais%7D%5C%3Af%5Cleft%28x%5Cright%29%5Cle%20%5C%3Ay_v)
![\mathrm{If}\:a>0\:\mathrm{the\:range\:is}\:f\left(x\right)\ge \:y_v](https://tex.z-dn.net/?f=%5Cmathrm%7BIf%7D%5C%3Aa%3E0%5C%3A%5Cmathrm%7Bthe%5C%3Arange%5C%3Ais%7D%5C%3Af%5Cleft%28x%5Cright%29%5Cge%20%5C%3Ay_v)
![a=1,\:\mathrm{Vertex}\:\left(x_v,\:y_v\right)=\left(0,\:0\right)](https://tex.z-dn.net/?f=a%3D1%2C%5C%3A%5Cmathrm%7BVertex%7D%5C%3A%5Cleft%28x_v%2C%5C%3Ay_v%5Cright%29%3D%5Cleft%280%2C%5C%3A0%5Cright%29)
![f\left(x\right)\ge \:0](https://tex.z-dn.net/?f=f%5Cleft%28x%5Cright%29%5Cge%20%5C%3A0)
Thus,
![\mathrm{Range\:of\:}x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\ge \:0\:\\ \:\mathrm{Interval\:Notation:}&\:[0,\:\infty \:)\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathrm%7BRange%5C%3Aof%5C%3A%7Dx%5E2%3A%5Cquad%20%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Af%5Cleft%28x%5Cright%29%5Cge%20%5C%3A0%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%5B0%2C%5C%3A%5Cinfty%20%5C%3A%29%5Cend%7Bbmatrix%7D)
The graph is also attached below.
Answer:
P = 4/13
Step-by-step explanation:
In a deck of 52 cards, there are 3 aces(spade, heart, diamond), 1 club ace, and 12 remaining club cards
=> The probability of randomly drawing 1 card that is an ace card or a club card:
P = number of elements/total number of elements
P = (3 + 1 + 12)/52
P = 16/52
P = 4/13
=> Option A is correct
So some quick tips for inequalities:
**What you do to one side you HAVE TO do it to the other 2**
**If you ever divide a negative number past the inequality signs, the signs FLIP!!**
(e.g. 2>-5x>25)
-2/5 <x < -5
I hope that this has helped!