Jan conducts an experiment and tosses a coin 40 times. She gets heads 30 times instead of the expected 20. What could be a reaso
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<em><u>This problem seems to be wrong because no minimum point was found and no point of landing exists</u></em>
Answer:
1) There is no maximum height
2) The ball will never land
Step-by-step explanation:
<u>Derivatives</u>
Sometimes we need to find the maximum or minimum value of a function in a given interval. The derivative is a very handy tool for this task. We only have to compute the first derivative f' and have it equal to 0. That will give us the critical points.
Then, compute the second derivative f'' and evaluate the critical points in there. The criteria establish that
If f''(a) is positive, then x=a is a minimum
If f''(a) is negative, then x=a is a maximum
1)
The function provided in the question is
Let's find the first derivative
solving h'=0:
x=3
Computing h''
h''(x)=4
It means that no matter the value of x, the second derivative is always positive, so x=3 is a minimum. The function doesn't have a local maximum or the ball will never reach a maximum height
2)
To find when will the ball land, we set h=0
Simplifying by 2
Completing squares
Factoring and rearranging
There is no real value of x to solve the above equation, so the ball will never land.
This problem seems to be wrong because no minimum point was found and no point of landing exists
Answer:
A = 20.5 sq units
Step-by-step explanation:
you have the height of 5 but need to figure out the base
we can use the Pythagorean Theorem two times to find it
1) a² + 5² = 7²
a² = 49-25
a = =
·
= 2
2) b² + 5² = 6²
b² = 36-25
b =
Base = 2 +
A = 1/2 x 5· (2 +
)
A = 2.5 · (2 +
)
A = 2.5(8.2)
A = 20.5
Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.
Step-by-step explanation:
To determine if solution exist or not, you test the equation for consistency.
A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.
Consider the following scenarios:
(1) For example:Given the matrix A=, to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:
Let A transpose be B.
∵B=
the system Bx=0 can be represented in matrix form as:
=
................................eq(1)
Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,
|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).
Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have :
=
. The rank of
can be found by using the second column and third column pair as follows:
||=(3*0)-(0*2)=0 i.e,
is a singular matrix with rank of order 1.
Note: a matrix is singular if its determinant is = 0 and non-singular if it is 0.
Comparing the rank of both B and , it is obvious that
Rank of BRank of
since (-2)<1.
Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution. </em>
(2) If B= is the transpose of matrix A=
, then
Then the equation Bx=0 is represented as:
=
..................................eq(2)
|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.
=
,
||=(5*0)-(0*10)=0-0=0 i.e
has a rank of order 1.
we can therefor conclude that since
rank B=rank =1, equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (
and
).
<u>Summary:</u>
- Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
- Determine the rank of both the coefficients matrix B and
which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
- If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.