Answer:
x= 15
Step-by-step explanation:
4x+15=75
4x = 60
x = 15
Answer:
32
Step-by-step explanation: OH NO MY SEXUALITY ITS BROCKEN
I: y<2*x
II: y<8
III: x>2
a: (5,3):
I: 3<2*5
3<10 ?yes
II: 3<8 ?yes
III: 5>2 ?yes
This is a solution
b: (6,0):
I: 0<2*6
0<12 ?yes
II: 0<8 ?yes
III: 6>2 ?yes
This is a solution
c: (3,3):
I: 3<2*3
3<6 ?yes
II: 3<8 ?yes
III: 3>2 ?yes
This is a solution
d: (1,-4):
I: -4<2*1
-4<2 ?yes
II: -4<8 ?yes
III: 1>2 ?no
This is not a solution
e: (4,-2):
I: -2<2*4
-2<8 ?yes
II: -2<8 ?yes
III: 4>2 ?yes
This is a solution
f: (5,9):
I: 9<2*5
9<10 ?yes
II: 9<8 ?no
This is not a solution
so a,b,c,e are solutions and d,f not
if the diameter of a circle is 15, its radius is half that or 7.5.
![\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=7.5 \end{cases} A=\pi (7.5)^2\implies A=56.25\pi \implies \stackrel{\pi =3.14}{A=176.625}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D7.5%20%5Cend%7Bcases%7D%20A%3D%5Cpi%20%287.5%29%5E2%5Cimplies%20A%3D56.25%5Cpi%20%5Cimplies%20%5Cstackrel%7B%5Cpi%20%3D3.14%7D%7BA%3D176.625%7D%20)
Consider the equation y = x^2. No matter what x happens to be, the result y will never be negative even if x is negative. Example: x = -3 leads to y = x^2 = (-3)^2 = 9 which is positive.
Since y is never negative, this means the inverse x = sqrt(y) has the right hand side never be negative. The entire curve of sqrt(x) is above the x axis except for the x intercept of course. Put another way, we cannot plug in a negative input into the square root function for this reason. This similar idea applies to any even index such as fourth roots or sixth roots.
Meanwhile, odd roots such as a cube root has its range extend from negative infinity to positive infinity. Why? Because y = x^3 can have a negative output. Going back to x = -3 we get y = x^3 = (-3)^3 = -27. So we can plug a negative value into the cube root to get some negative output. We can get any output we want, negative or positive. So the range of any radical with an odd index is effectively the set of all real numbers. Visually this produces graphs that have parts on both sides of the x axis.
