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Tcecarenko [31]
3 years ago
9

HELP:. In baseball, the distance from the pitcher's mound to the batter is 60.5 feet. A pitcher can throw the baseball at 121 fe

et per second (82.5 miles per hour). An eye blink is 0.1 second. In how many blinks of an eye does it take for the baseball to travel from the pitcher to the batter?
Mathematics
1 answer:
Ray Of Light [21]3 years ago
3 0

Answer:

Hello the answer to your question is 0.4 sec.

Step-by-step explanation:

Hope this helps

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(c) One lorry travels from your town to another town. The lorry reaches a top speed
Katen [24]

Answer:

The average speed is 24 km/h less than the top speed of the lorry.

Step-by-step explanation:

The top speed of the lorry is the highest speed reached during transit, while its average speed is the mean speed attained

Given:.

Top speed of the lorry = 90 km/h

Average speed of the lorry = 66 km/h

Then,

The difference between the speed reached = top speed - average speed

                                            = 90 - 66

                                            = 24 km/h

Therefore, the average speed is 24 km/h less than the top speed of the lorry.

8 0
3 years ago
Shadow Software company earned a total of 800,009 selling programs during the year 2012. 125,300 of taht amount was used to pay
Softa [21]
Subtract 125,300 from 800,009. From there you find that they made 674,709 in profit. Answer is 674,709
6 0
3 years ago
Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the
Zarrin [17]

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=\left[\begin{array}{ccc}1&2\\3&4\end{array}\right], 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=\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]

the system Bx=0 can be represented in matrix form as:

\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right] ................................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 B_{0}:      

B_{0}=\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]. The rank of B_{0} can be found by using the second column and third column pair as follows:

|B_{0}|=(3*0)-(0*2)=0 i.e, B_{0} 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 \neq0.

Comparing the rank of both B and B_{0}, it is obvious that

Rank of B\neqRank of B_{0} since (-2)<1.

Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution.     </em>

(2) If B=\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right] is the transpose of matrix A=\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right], then

Then the equation Bx=0 is represented as:

\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]=\left[\begin{array}{ccc}0\\0\end{array}\right]..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0  i.e B has a rank of order 1.

B_{0}=\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right],

|B_{0}|=(5*0)-(0*10)=0-0=0   i.e B_{0} has a rank of order 1.

we can therefor conclude that since

rank B=rank B_{0}=1,  equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (X_{1} and X_{2}).

<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 B_{0} 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.
5 0
3 years ago
NEED HELP ASAP PLEASE WITH #9 and #10!!
ICE Princess25 [194]
9) Because the total number of pencils is 180 and you will use them up in 30 days, the equation will have to equal 0 total pencils when 30 is substituted in for the time factor, or x. This already takes our choice 3 because it doesn’t meet this criteria.
The answer to how many pencils in 20 days could be answered by plugging in 20 for x. Choice 4 cannot work because it results in a negative number of pencils. Choices 1 and 2 use the same equation, so by plugging in 20 it is clear choice 2 is the correct answer.

10) A line parallel to the go en equation would have the same slope, -3, which means choices 1 and 4 are out. Plug in (-2,5) into both choices 2 and 3. Plugging -2 into x in choice 2 gives -5, and in choice 3 gives 5 for the y value. Therefore choice 3 is correct.
8 0
3 years ago
3x^5+6x^3<br><br> PRIME<br> b<br> 3x^4(3x + 2)<br> 3(x^5 + x^3)<br> d<br> 3x^3(x^2 + 2x)
Gelneren [198K]

Answer:

Im guessing 33(2+2)

8 0
2 years ago
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