This is the concept of application of quadratic expressions. Given that the height of the ball is modeled by the equation;
h=-7.3t^2+8.25t+2.1+5
The time taken for the ball to hit the ground will be given as falls;
-7.3t^2+8.25t+7.1=0
to solve for t we use the quadratic formula;
t=[-b+/-sqrt(b^2-4ac)]/(2a)
a=-7.3, b=8.25, c=2.1
t=[-8.25+/-sqrt[8.25^2+4*7.3*7.1]/(-2*7.3)
t= -0.572
or
t=1.702
since there is not negative time we take the time taken for the ball to hit the ground will be: t=1.702 sec
Answer:
The answer to your question is x<-8, hope this helps :)
Answer:
3) (2,-9)
4) (0,-5)
5) (1,-8)
Step-by-step explanation:
3)
The vertex will occur between you x-intercepts.
You already found that happens at x=2.
To find the corresponding y-coordinate, replace x in
f(x)=(x+1)(x-5) with 2:
f(2)=(2+1)(2-5)
f(2)=(3)(-3)
f(2)=-9
So the vertex is (2,-9).
4)
The y-intercept is when x=0.
So in f(x)=(x+1)(x-5) replace x with 0:
f(0)=(0+1)(0-5)
f(0)=(1)(-5)
f(0)=-5
So the y-intercept is (0,-5).
5)
To find another point just plug in anything besides any x already used.
We preferably want to use a value of x that will keep us on their grid however far up,down,left, or right their grid goes out. So I'm going to choose something close to the vertex which is at x=2. Let's go with x=1.
So replace x in f(x)=(x+1)(x-5) with x=1:
f(1)=(1+1)(1-5)
f(1)=(2)(-4)
f(1)=-8
So another point to graph is (1,-8).
The answer of this question is 152
Answer:
g = 1
Step-by-step explanation:
-3 + 5 + 6g = 11 - 3g
2 + 6g = 11 - 3g
Add 3g to both sides.
2 + 9g = 11
Subtract 2 from both sides
9g = 9
g = 1
