Answer: The equation is y = -6*x
Step-by-step explanation:
I suppose that we want to find the equation for a line that passes through the point (-1, 6) and the origin (remember that the origin is the point (0,0))
A general linear equation is written as:
y = a*x + b
Where a is the slope and b is the y-intercept.
If this line passes through the points (x₁, y₁) and (x₂, y₂), then the slope of the line is equal to:
a = (y₂ - y₁)/(x₂ - x₁)
Now we know that our line passes through the points (0, 0) and (-1, 6), then the slope is:
a = (6 - 0)/(-1 - 0) = 6/-1 = -6
Then our equation is something like:
y = -6*x + b
To find the value of b we can use the fact that this line passes through the point (0, 0).
This means that when x = 0, y is also equal to zero.
If we replace these values in the equation we get:
0 = -6*0 + b
0 = b
Then our equation is:
y = -6*x
Answer:
The measure of ∠EFG is 52°
Step-by-step explanation:
Given line m is parallel to line p. m∠HEF = 39º and m∠IGF = 13º.we have to find m∠EFG.
In ΔJFG,
By angle sum property of triangle, which states that sum of all angles of triangle is 180°
m∠FJG+m∠JGF+m∠JFG=180°
⇒ 39°+13°+m∠JFG=180°
⇒ m∠JFG=180°-39°-13°=128°
As JFE is a straight line ∴ ∠JFG and ∠EFG forms linear pair
⇒ m∠JFG+m∠EFG=180°
⇒ 128°+m∠EFG=180°
⇒ m∠EFG=52°
The measure of ∠EFG is 52°
Answer:
Step-by-step explanation:
Since radar reports 30 tracks in one minute in which 60 seconds makes 1 minute
Let calculate using this formula
Using this formula
Percentage of time=Tracks reported in minutes/60 seconds*100
Let plug in the formula
Percentage of time=30 tracks/60 seconds*100
Percentage of time=0.5*100
Percentage of time=50%
Inconclusion The percentage of the time that radar track the aircraft is 50%
Three hundred forty eight thousand five hundred.
Hope this helps!
Answer:
Step-by-step explanation:
Let the no of cars be x and no of buses be y.
x + y = 60 -> y = 60 - x
6x + 30y = 600
with the available equations let us calculate the values of x and y.
6x + 30 * (60-x) = 600
6x + 1800 - 30x = 600
24x = 1200
x cars = 50
y buses = 10
maximum profit = 50 * 2.5 + 10 * 7.5 = $200
