Apparently I'm supposed to do one at a time so here we go again

Ben says that n=5 is the solution of the equation 7n=45. How can you check whether Ben is correct?

GalinKa [24]

Answer:

Its not. 45 would have to be 35 for n to be 5.

Step-by-step explanation:

7(5)=45

3 0

2 years ago

Which matrix represents the system of equations shown below? 3x-5y=12 4x-2y=15

tatuchka [14]

Answer:

$\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]$

Step-by-step explanation:

When making a matrix of two equations with the variables x and y, the result will be a matrix with three columns:

a column for the values of x in each equation
a column for the values of y in each equation
a column for the independent values of each equation

since our system of equations is:

$3x-5y=12\\ 4x-2y=15$

we can see that the value for x in the first equation is 3 and in the second equation is 4, thus the first column will have the numbers 3 and 4:

$\left[\begin{array}{ccc}3&&\\4&&\\\end{array}\right]$

Now for the values of y we hvae -5 in the first equation and -2 in the second equation, we update the matrix with another column with the values of -5 and -2:

$\left[\begin{array}{ccc}3&-5&\\4&-2&\\\end{array}\right]$

Finally, the last column is the independent values of each equation (or the results) in the first equation that number is 12 and in the second equation is 15, thus the matrix is:

$\left[\begin{array}{ccc}3&-5&12\\4&-2&15\\\end{array}\right]$

usually there is a line separating the columns for the values of x and y, and the independent values:

$\left[\begin{array}{ccc}3&-5 &|12\\4&-2 &|15\\\end{array}\right]$

this is the matrix of the system of equations

6 0

2 years ago

A college library has five copies of a certain text on reserve. Three copies (1, 2, and 3) are first printings, and the remainin

dezoksy [38]

Answer:

S = { (4), (5), (1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (1,2,3,4), (1,2,3,5), (2,1,3,4) (2,1,3,5), (3,1,2,4), (3,1,2,5) }

A ={(4), (5)}

Step-by-step explanation:

Given that:

Among the three copies, (1,2,3) are the first printings, and (4,5) are the second printings.

A student who examines these books in random order stops when a second printing has been selected.

Thus, we can compute the sample space associated with these experiments as:

S = { (4), (5), (1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (1,2,3,4), (1,2,3,5), (2,1,3,4) (2,1,3,5), (3,1,2,4), (3,1,2,5) }

Suppose A represents the event that we must examine exactly one book.

Then the outcomes of A are:

A ={(4), (5)}

6 0

2 years ago

Divid 15 into two parts such that the sum of their reciprocal is 3/10

emmasim [6.3K]

Let x and y be the 2 parts of 15 ==> x + y=15 (given)

Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)

Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)

==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy

But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50

Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5

4 0

2 years ago

Hello please help i’ll give brainliest

Delvig [45]

Answer:

the bin 60-69 contain the most data values

3 0

2 years ago

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