Answer:
Step-by-step explanation:
Given
Parabola has x-intercept has 
and 
and Y-intercept as 
Now the general equation of parabola is
Substitute in 
we get
Now substitute in equation
Now substitute in equation
Solving 
and 
we get
therefore
Y= 4
How to:
Using the y axis it hits the line at 4
Answer:
y = 11(x -1) +6
Step-by-step explanation:
The slope at the given point is ...
y' = 14x -3x^2 = 14(1) -3(1^2) = 11
Then the point-slope equation for the line can be written as ...
y = m(x -h) +k . . . . . . for slope m at point (h, k)
y = 11(x -1) +6
Subtract 3 from both sides so that the equation becomes -2x^2 + 5x - 13 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
x = [ -b ± √(b^2 - 4ac) ] / (2a)
x = [ -5 ± √((5)^2 - 4(-2)(-13)) ] / ( 2(-2) )
x = [-5 ± √(25 - (104) ) ] / ( -4 )
x = [-5 ± √(-79) ] / ( -4)
Since √-79 is nonreal, the answer to this question is that there are no real solutions.
Answer:
find the area if the sector,then the area of the triangle.After that, Subtract the area of the triangle from the area of the sector
