<u>Answer:</u>
Hence, Relation t is a function. The inverse of relation t is a function.
<u>Step-by-step explanation:</u>
We are given the relation as:
x: 0 , 2 , 4 , 6
y: -10 , -1 , 4 , 8
<em>Clearly from the y-values corresponding to the x-values we could see that each x has a single image (single y-value).</em>
Hence, the corresponding relation is a function.
Now we have to find whether the inverse of this relation is a function or not.
When we take the inverse of this function that is the y-values will behave as a pre-image and x-values as its image.
Hence we will see that corresponding to each y-value there is a unique image hence the inverse relation is also a function.
Hence, Relation t is a function. The inverse of relation t is a function.
Answer:
36 erasers
Step-by-step explanation:
Let number of erasers be e
let number of rulers be r
We can write:
e + r = 70
and
After giving away, he has
Erasers: 2/3e
Rulers: r - 10
These two are equal, so we can write and solve:
2/3e = r - 10
2/3e + 10 = r
Putting this in initial equation, we have:
e + (2/3e + 10) = 70
5/3e + 10 = 70
5/3 e = 60
e = 36
And rulers is:
r = 2/3(36) + 10 = 34
Hence, he had 36 erasers in the beginning
So, we know the center is at -1, -3, hmmm what's the radius anyway?
well, the radius will be the distance from the center to any point on the circle, it just so happen that we know -7, -5 is on it, thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -1 &,& -3~) % (c,d) &&(~ -7 &,& -5~) \end{array} \\\\\\ d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ r=\sqrt{[-7-(-1)]^2+[-5-(-3)]^2}\implies r=\sqrt{(-7+1)^2+(-5+3)^2} \\\\\\ r=\sqrt{36+4}\implies r=\sqrt{40}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-1%20%26%2C%26%20-3~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%20-7%20%26%2C%26%20-5~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%5B-7-%28-1%29%5D%5E2%2B%5B-5-%28-3%29%5D%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B%28-7%2B1%29%5E2%2B%28-5%2B3%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B36%2B4%7D%5Cimplies%20r%3D%5Csqrt%7B40%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-1}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\sqrt{40}}{ r} \\\\\\\ [x-(-1)]^2+[y-(-3)]^2=(\sqrt{40})^2\implies (x+1)^2+(y+3)^2=40](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%20%0Acenter~~%28%5Cstackrel%7B-1%7D%7B%20h%7D%2C%5Cstackrel%7B-3%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20%0Aradius%3D%5Cstackrel%7B%5Csqrt%7B40%7D%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-1%29%5D%5E2%2B%5By-%28-3%29%5D%5E2%3D%28%5Csqrt%7B40%7D%29%5E2%5Cimplies%20%28x%2B1%29%5E2%2B%28y%2B3%29%5E2%3D40)
Inequalities are used to represent unequal expressions
The solution to the inequality is: 
<h3>How to determine the inequality solution</h3>
The inequality is given as:
60>5(6-2x)
Rewrite the inequality properly as:
Divide both sides of the inequality by 5
Subtract 6 from both sides of the inequality
Divide both sides of the inequality by -2
![-3Rewrite the inequality as[tex]x > -3](https://tex.z-dn.net/?f=-3%3C%2Fp%3E%3Cp%3ERewrite%20the%20%3Cstrong%3Einequality%3C%2Fstrong%3E%20as%3C%2Fp%3E%3Cp%3E%5Btex%5Dx%20%3E%20-3)
Hence, the solution to the inequality is:
Read more about inequalities at:
brainly.com/question/11234618
