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alexandr402 [8]
3 years ago
9

I) Domain, ii) x-intercept, iii) y-intercept, iv) vertical asymptote, and v) horizontal asymptote of

Mathematics
1 answer:
Montano1993 [528]3 years ago
7 0

Answer:

i)

Step-by-step explanation:

i) you can't divide by 0 and when you put x=5 into the bottom part 4x-20 = 4(5)-20 = 20-20 = 0 so x cannot be 5. So the domain is x > 5 and x < 5 OR in other words, when x is not 5.

ii) substitute y=0 into the equation and solve the quadratic on top

iii) substitute x=0 into the equation and write down the value

horizontal asymptote at x=5 because we found out that x cannot be 5 in part i)

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Answer:

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Step-by-step explanation:

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NEED HELP QUICK!! :)
Vlada [557]
Let x and y be your two consecutive whole numbers

x < sqrt(142) < y
x^2 < 142 < y

So, we are looking for x and y such that x^2 < 142 and y^2 > 142.

The closest squared number to 142 is 144 = 12^2.
Next is 11^2 = 121.

11 and 12 are consecutive.

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3 years ago
(2x + 8) (3x + 17) solve for x
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Step-by-step explanation:

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3 years ago
Write an equation in slope-intercept form of the line that passes through (-5, 19) and (5.13).
ruslelena [56]

Answer:

y=-3/5x+16

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(13-19)/(5-(-5))

m=-6/(5+5)

m=-6/10

simplify

m=-3/5

y-y1=m(x-x1)

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3 0
2 years ago
[10] In the following given system, determine a matrix A and vector b so that the system can be represented as a matrix equation
irina1246 [14]

Answer:

y=-\frac{158}{579}

Step-by-step explanation:

To find the matrix A, took all the numeric coefficient of the variables, the first column is for x, the second column for y, the third column for z and the last column for w:

A=\left[\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right]

And the vector B is formed with the solution of each equation of the system:b=\left[\begin{array}{c}3\\-3\\6\\1\end{array}\right]

To apply the Cramer's rule, take the matrix A and replace the column assigned to the variable that you need to solve with the vector b, in this case, that would be the second column. This new matrix is going to be called A_{2}.

A_{2}=\left[\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right]

The value of y using Cramer's rule is:

y=\frac{det(A_{2}) }{det(A)}

Find the value of the determinant of each matrix, and divide:

y==\frac{\left|\begin{array}{cccc}1&3&2&2\\-7&-3&5&-8\\4&6&1&1\\3&1&-1&1\end{array}\right|}{\left|\begin{array}{cccc}1&1&2&2\\-7&-3&5&-8\\4&1&1&1\\3&7&-1&1\end{array}\right|} =\frac{158}{-579}

y=-\frac{158}{579}

7 0
3 years ago
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