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2 years ago

3. Consider the straight line x + y = 1 and the family of straight lines that pass

Mathematics

1 answer:

ANTONII [103]2 years ago

5 0

Answer:

solution: (1/(1+m), m/(1+m))
no solution for m = -1 (lines are parallel)

Step-by-step explanation:

Two distinct lines will always have a point of intersection, provided they are not parallel. That is, the system of equations {x +y = 1, y = mx} will have a unique solution as long as m ≠ -1.

The solution to the system can be found by substitution:

x +mx = 1

x(1 +m) = 1 . . . . factor out x

x = 1/(1+m) . . . . . divide by the coefficient of x; we must have m ≠ -1

y = m/(1+m) . . . . find the y-coordinate of the solution

The point of intersection is (1/(1+m), m/(1+m)).

The above solution is undefined when the denominator is zero: m = -1. There is no solution for m = -1.

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A rectangular prism has a base with a length of 25, a width of 9, and a height of 12. A second prism has a square base with a si

RSB [31]

Answer:

The height of the second prism is 12 units.

Step-by-step explanation:

First, we must know that the volume of a rectangular prism is equal to length x width x height. With this information, we can multiply the given dimensions of the first rectangular prism to get a volume of 2700 cubic units. From here, we must understand wha we need from the second rectangular prism. Because it has a square base of 15, that accounts for both its width and length. Therefore, we have that

$Volume= 15^2 x\\2700/225=x\\12=x$

6 0

2 years ago

Each point on the coordinate plane has an address called

RideAnS [48]

I do not understand what to do , more explicitly sorry :(

8 0

3 years ago

Express the system as AX = B; then solve using matrix inverses

Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]$

$\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B$

Adj.A=Transpose of cofactor of Matrix A

$Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2$

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]$

$\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4$

7 0

3 years ago

If n is a term of the sequence 2,5,8,11,..., which expression would give the value of n?

Wittaler [7]

The term of the sequence is eleven 11

5 0

3 years ago

Read 2 more answers

A city grid of Anytown, USA is shown on the grid below. The fire department is represented by quadrilateral RSTU. Another fire d

Alekssandra [29.7K]

Answer: option d. C (0,3), D (0,5).

Justification:

1) The x - coordinates of the vertices A and B are shown in the diagrama, They are both - 4, so the new vertices C and D must be in a line parallel to y = - 4.

2) The y-coordinates of the vertices A and B are also shown in the diagrama. They are equal to 3 and 5 respectively.

3) We can see that the new points C and D must be over a parallel line to y = - 4 and that their distance to the points A and B has to be the same distance of the point R and S to U and T.

That distance is 4, so the line may be y = - 7 or y = 0.

4) If the line is y = 7 the points C and D would have coordinates (-7,3) and (-7,5), but this points are not among the options.

5) If the line is y = 0 the points C and D would have coordinates (0, 3) and (0,5), which is precisely the points of the option d. That is the answer.

3 0

3 years ago

Read 2 more answers