Answer:
Let X be 2
Now,
y = 5*2 - 8
or, y= 10-8
or, y =2
Hope this will help please mark me as <em>brainlest.</em>
Answer:
It is a many-to-one relation
Step-by-step explanation:
Given
See attachment for relation
Required
What type of function is it?
The relation can be represented as:
![\left[\begin{array}{c}x\\ \\-5\\-2\\1\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5C%20%5C%5C-5%5C%5C-2%5C%5C1%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}y\\ \\10\\11\\4\\10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy%5C%5C%20%5C%5C10%5C%5C11%5C%5C4%5C%5C10%5Cend%7Barray%7D%5Cright%5D)
Where
and 
Notice that the range has an occurrence of 10 (twice)
i.e.
and 
In function and relations, when two different values in the domain point to the same value in the range implies that, <em>the relation is many to one.</em>
These problems are called systems of equations. Basically you have two linear equations and you need to find the values for x and y. In other words, all these equation are lines and our answer will be the exact point that the pair of lines intersect. For example, if we get x=1 and y=2 the lines will intersect at point (1,2). Now that you have some background knowledge here comes the tricks and tactics kid.
We know that we can solve one variable equation easily. For example...
x+1=2
x=1 obviously
Cause we have two variables x and y it is not possible to find a solution. For example, in the equation x+y=10, x=1 when y=9 and x=2 when y=8. There is not correct answer.
So what can we do? We have to make a two variable equation into a one variable equation.
There are two ways to do this: substitution and elimination. I will create a sample problem and then solve it using both methods.
x+y=2
2y-y=1
3)
-3x-5y=-7 -----> -12x-20y=-28
-4x-3y=-2 ------> -12x-9y=-6
-12x-20y=-28
-(-12x-9y=-6)
---------------------
-11y=-22
y=2
-3x-5(2)=-7
-3x=3
x=-1
4) 8x+4y=12 ---> 24x+12y=36
7x+3y=10 ---> 28x+12y=40
28x+12y=40
-(24x+12y=36)
---------------------
4x=4
x=1
8(1)+4y=12
4y=4
y=1
5) 4x+3y=-7
-2x-5y=7 ----> -4x-10y=14
4x+3y=-7
+(-4x-10y=14)
-------------------
-7y=7
y=-1
4x+3(-1)=-7
4x=-4
x=-1
6) 8x-3y=-9 ---> 32x-12y=-36
5x+4y=12 ---> 15x+12y=36
32x-12y=-36
+(15x+12y=36)
--------------------
47x=0
x=0
8(0)-3y=-9
-3y=-9
y=3
7)-3x+5y=-2
2x-2y=1 ---> x-y=1/2 ----> x=y+1/2
-3(y+1/2)+5y=-2
-3y-1.5+5y=-2
2y=-0.5
y=0.25
2x-2(0.25)=1
2x=1.5
x=0.75
Answer:
you need 5 buses and 4 will be totally full and one will only be 1/3 full
or you can have 14% of all 5 buses with empty seats
Step-by-step explanation:
The answer is would have to be b.
