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3 years ago

3x+2y=6 and y=1/3x-11 perpendicular or parallel

1 answer:

6 0

Answer: To determine if two lines are parallel or perpendicular, you have to look at the slopes. For the line y=kx+b, k is the slope.

For two lines y=k1x+b1 and y=k2X+b2 ,When K1=K2, two lines are parallel; when K1=-1/k2, two lines are perpendicular. It doesn't matter whatever b1, b2 are. Hence to save time, you only need to calculate k1 and k2.

To calculate the slope k of any line, you have to change the equation to y=kx+b

For -y=3x-2, k=3/(-1)=-3 (you divide -1 on each side of =, but you don't need to calculate -2/(-1))

For -6X+2y=6, K=6/2=3 (you first move the -6x to right side, it becomes 6x, then divide by 2)

Now you can get the answer: The two lines neither parallel nor perpendicular.

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The University of Texas at Austin has three times as many students enrolledas the University of Miami. The University of Califor

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University of Texas has 46,500 students, University of California has 34,000 students and University of Miami has 15,500 students.

Step-by-step explanation:

Represent Texas with T, Miami with M and California with C. With the given information, the following equations were obtained.

T = 3 × M

T – 3M = 0 ………………………….(i)

C = 3,000 + 2 × M

C – 2M = 3,000…………………..(ii)

T + C + M = 96,000…………….(iii)

By rearranging, we have:

1T + 0C – 3M = 0 ………………………….(i)

0T + 1C – 2M = 3,000…………………..(ii)

1T + 1C + 1M = 96,000………….…….(iii)

The Augmented Matrix is:

$\left[\begin{array}{ccc}1&0&-3\\0&1&-2\\1&1&1\end{array}\right]$ $\left[\begin{array}{ccc}T\\C\\M\end{array}\right]$ = $\left[\begin{array}{ccc}0\\3,000\\96,000\end{array}\right]$

Making $\left[\begin{array}{ccc}T\\C\\M\end{array}\right]$ the subject of the formula we have:

$\left[\begin{array}{ccc}T\\C\\M\end{array}\right]$ = $\frac{\left[\begin{array}{ccc}0\\3,000\\96,000\end{array}\right]}{\left[\begin{array}{ccc}1&0&-3\\0&1&-2\\1&1&1\end{array}\right]}$

Let’s say matrix A = $\left[\begin{array}{ccc}1&0&-3\\0&1&-2\\1&1&1\end{array}\right]$ , the inverse of matrix A = $\left[\begin{array}{ccc}\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} \\-\frac{1}{3} &\frac{2}{3} &\frac{1}{3} \\-\frac{1}{6} &-\frac{1}{6} &\frac{1}{6} \end{array}\right]$ is determined by the formula $\frac{Adjoint(A)}{determinant(A)}$ . This can also be computed directly on a scientific calculator.

$\left[\begin{array}{ccc}T\\C\\M\end{array}\right]$ = $\left[\begin{array}{ccc}\frac{1}{2} &-\frac{1}{2} &\frac{1}{2} \\-\frac{1}{3} &\frac{2}{3} &\frac{1}{3} \\-\frac{1}{6} &-\frac{1}{6} &\frac{1}{6} \end{array}\right]$ $\left[\begin{array}{ccc}0\\3,000\\96,000\end{array}\right]$

= $\left[\begin{array}{ccc}46,500\\34,000\\15,500\end{array}\right]$

T = 46,500

C = 34,000

M = 15,500

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4 years ago

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