Answer:
A
Step-by-step explanation:
Answer:
Second point (-5/2, -7/2)
First point (3/2, 17/2)
Step-by-step explanation:
We have two equations, and we want to know at wich poin are equal. Hence, we have a system of equations and the solution is nothing more that the point (x,y) where those functions intercepts.
4x2+ 7x -11=y
3x+4=y
Lets use substitute method
4x2+7x-11=3x+4
This can be re arrange as the following eq:
4x2+4x-15=0
A quadratic equation, its solution can be obtained using the below eq.

where a=4, b=4, c=-15.
Remember, the quadratic equation as a +/- sign, meaning that you will obtain one answer using the + operator and other using the - operator.
By doing the above, we have x=-5/2 and x=3/2
By using x=3/2 in equation of line (3x+4=y) we have y=17/2
First point (3/2, 17/2)
By using x=-5/2 in equation of line (3x+4=y) we have y= -7/2
Second point (-5/2, -7/2)
Those points are the ones where the line and the parabola intercept.
Answer:
i cant see it
Step-by-step explanation:
Answer:
Value of 
Step-by-step explanation:
The given triangle is isosceles
Here

For this case we have that the equation that describes the height is:

Where,
v: initial speed
c: initial height
Substituting values we have:

Then, to find the time, we use the quadratic formula:

Substituting values we have:

Rewriting:



We discard the negative root because we want to find the time.
We have then:


Answer:
D. 