X=11 because u Solve for the first variable in one of the equations, then substitute the result into the other equation.
Considering High School level question, answer can be written as:
A system of 2 linear equations is [two] dimensional. It is a graph of [two] lines. The solutions can be [unique] solution if the graph intersects. [No] solution if the lines are parallel - meaning they have the same slope, or [Infinitely many] solutions if they are the same line.
Explanation:
when two lines are drawn on a two-dimensional plane then there are only three possible cases:
Case1: lines will intersect
In that case you will get a unique solution at the intersection point.
Case2: lines are parallel but don't touch each other
In that case there will be no point which lies on both lines so No solution.
Case3: lines are overlapping.
In that case all the points lies on both lines so infinitely many solutions.
Answer:
Step-by-step explanation:
The equation of line having slope 
and passing through 
.
Answer:
Given the function f(x) = 3x + 1, evaluation of f(a + 1) gives:
C. 3a + 4
Step-by-step explanation:
Given function:
f(x) = 3x + 1
We have to find f(a+1).
For this purpose, we will take x = a+1 and
substitute it in the function f(x) = 3x+1:
f(x) = 3x + 1
f(a+1) = 3(a+1) +1
f(a+1) = 3(a) + 3(1) +1
f(a+1) = 3a+3+1
f(a+1) = 3a + 4
So the function f(a+1) is equal to option C. 3a + 4.
Answer:
Step-by-step explanation:
We have been given that an arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up be the positive direction. Because gravity is the force pulling the arrow down, the initial acceleration of the arrow is −9.8 meters per second squared.
We know that equation of an object's height t seconds after the launch is in form 
, where
g = Force of gravity,
= Initial velocity,
= Initial height.
For our given scenario 
, 
and 
. Upon substituting these values in object's height function, we will get:
Therefore, the function for the height of the arrow would be .
