Answer:
a = 2, b = -9, c = 3
Step-by-step explanation:
Replacing x, y values of the points in the equation y = a*x^2 + b*x +c give the following:
(-1,14)
14 = a*(-1)^2 + b*(-1) + c
(2,-7)
-7 = a*2^2 + b*2 + c
(5, 8)
8 = a*5^2 + b*5 + c
Rearranging:
a - b + c = 14
4*a + 2*b + c = -7
25*a + 5*b + c = 8
This is a linear system of equations with 3 equations and 3 unknows. In matrix notation the system is A*x = b whith:
A =
1 -1 1
4 2 1
25 5 1
x =
a
b
c
b =
14
-7
8
Solving A*x = b gives x = Inv(A)*b, where Inv(A) is the inverse matrix of A. From calculation software (I used Excel) you get:
inv(A) =
0.055555556 -0.111111111 0.055555556
-0.388888889 0.444444444 -0.055555556
0.555555556 0.555555556 -0.111111111
inv(A)*b
2
-9
3
So, a = 2, b = -9, c = 3
Answer:
A. square root of a^2 + b^2 for both answers
Step-by-step explanation:
The first problem, we are given
a^2 + b^2 = c^2
What we do is solve for c.
sqrt(a^2 + b^2) = c
c = sqrt(a^2 + b^2)
For problem 2,
WE can apply the Pythagorean theorem because we have a right triangle.
The equations is
a^2 + b^2 = c^2 like the first problem
Solving gets us
sqrt(a^2 + b^2) = c
c = sqrt(a^2 + b^2)
Answer:
ABD=180 and EHF=90
Step-by-step explanation:
Answer:
See explanation
Step-by-step explanation:
cos(θ)= 18/30 = 3/5
sin(θ)= 24/30 = 4/5
tan(θ)= 24/18 = 4/3
cos(ϕ)= 24/30 = 4/5
sin(ϕ)= 18/30 = 3/5
tan(ϕ)= 18/24 = 3/4
Answer:
Hence, Grasshopper will land on the ground after 1.5 sec.
Step-by-step explanation:
It s given that:
The height, in feet, of the grasshopper above the ground after t seconds is modeled by the function:
Now we are asked to find:
In how many seconds will the grasshopper land on the ground?
i.e. we have to find the value of t such that h(t)=0
i.e.
i.e. we need to find the roots of the given quadratic equation.
On solving the quadratic equation or plotting it's graph we could observe that the point where h(t)=0 are:
As time can't be negative hence we will consider:
Hence, grasshopper will land on the ground after 1.5 sec.
