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3 years ago

What I the slope of a line that is perpendicular to the line 2y-3x=8

Mathematics

1 answer:

Zolol [24]3 years ago

6 0

ANSWER

$- \frac{2}{3}$

EXPLANATION

The given given equation is

$2y - 3x = 8$

We need to rewrite this equation in the slope-intercept form:

$y = mx + b$

We add 3x to both sides.

$2y - 3x + 3x=8 + 3x$

$\implies \: 2y = 3x + 8$

We divide through by 2 to get,

$y = \frac{3}{2}x + 4$

The slope of this line is

$m = \frac{3}{2}$

Let the slope of the line perpendicular to this line be 'n' .

Then the product of the slopes of two perpendicular lines is always negative 1.

$m \times n = - 1$

$\implies \: \frac{3}{2} n = - 1$

$\implies \: \frac{2}{3} \times \frac{3}{2}n = - 1 \times \frac{2}{3}$

$n = - \frac{2}{3}$

Therefore the slope of the new line is

$- \frac{2}{3}$

You might be interested in

Help me and I’ll give brainliest! Plsss

olganol [36]

Answer:
(#14 - supplementary) x = 10.5
(#15 - supplementary) x = 10

Explanation:
For these problems, we must know that a supplementary angle can be viewed as a sum of angles that add to 180 degrees. With this known, we can write our equations to find x.

(#14)
15x - 12 + 5x - 18 = 180
20x - 30 = 180
20x = 210
x = 10.5

(#15)
6x + 13 + 14x - 33 = 180
20x - 20 = 180
20x = 200
x = 10

Hope this helps.

Cheers.

4 0

2 years ago

Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you

adell [148]

Answer:

1) The solution of the system is

$\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) The solution of the system is

$\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

Step-by-step explanation:

1) To solve the system of equations

$\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right$

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

$\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 2: Swap rows 1 and 2

$\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 3: $\left(R_1=\frac{R_1}{6}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 5: $\left(R_2=\frac{R_2}{3}\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]$

Step 6: $\left(R_3=R_3+R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]$

Step 7: $\left(R_3=\left(\frac{6}{37}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 8: $\left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 9: $\left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]$

Step 10: Rewrite the system using the row reduced matrix:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right$

2) To solve the system of equations

$\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right$

using the row reduction method you must:

Step 1:

$\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 2: $\left(R_1=\frac{R_1}{4}\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 3: $\left(R_2=R_2-\left(2\right)R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]$

Step 4: $\left(R_3=R_3-R_1\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 5: $\left(R_2=\left(2\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 6: $\left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]$

Step 7: $\left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)$

$\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]$

Step 8: $\left(R_3=\left(- \frac{2}{117}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 9: $\left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 10: $\left(R_2=R_2-\left(15\right)R_3\right)$

$\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]$

Step 11:

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right$

8 0

3 years ago

Which is bigger 3/8 or 0.3

Luba_88 [7]

$\frac{3}{8} \\
0.3=\frac{3}{10}$

If two fractions have the same numerator, the fraction with the smaller denominator is bigger.
$8\ \textless \ 10 \Rightarrow \frac{3}{8} \ \textgreater \ \frac{3}{10}$

3/8 is bigger.

3 0

3 years ago

Read 2 more answers

Calcula la longitud en YARDAS de

tatiyna

Answer:

Las longitudes solicitadas en yardas son:

Trayecto A = 109.361 yardas.
Trayecto B = 20.231785 yardas.

Step-by-step explanation:

Para hacer la conversión de unidades que requieres en el ejercicio, debes saber que:

1 metro = 1.09361 yardas

Con ese factor de conversión tú puedes hacer reglas de tres para calcular las medidas que requieres. En el caso del trayecto A:

Si:

1 metro = 1.09361 yardas
100 metros = X

Entonces:

$x=\frac{100m *1.09361yardas}{1m}$

Cancelamos metros y obtenemos:

x = 100 * 1.09361 yardas
x = 109.361 yardas

En este caso, el trayecto A en yardas corresponde a 109.361 yardas. El mismo procedimiento puede aplicarse para el trayecto B:

Si:

1 metro = 1.09361 yardas
18.50 metros = X

Entonces:

$x = \frac{18.50m*1.09361yardas}{1m}$

Cuando se cancelan los metros se obtiene:

x = 18.50 * 1.09361 yardas
x = 20.231785 yardas

Así, el trayecto B en yardas corresponde a 20.231785 yardas.

4 0

3 years ago

A. A group of twelve art students are visiting a local art museum for a field trip. The total cost of admission for the students

Shalnov [3]

Answer:

a) $10.42

b) 11 trips

c) 10 cookies

d) 12/125ths of a cheesecake (or 9.6%)

Step-by-step explanation:

Let x = the cost of admission for each student

Let y = the total cost of admission

⇒ y = 12x

⇒ 125 = 12x

⇒ x = 125 ÷ 12

⇒ x = 10.42

So the cost of admission for each student = $10.42

Let x = the number of trips

Let y = the total number of passengers

⇒ y = 12x

⇒ 125 = 12x

⇒ x = 125 ÷ 12

⇒ x = 10.42

Therefore, the van must make 11 trips in order to bring 125 passengers to the same location. (If it took 10 trips, then it would only bring 120 passengers, hence why we have to round make 11 trips to bring the last 5 passengers).

Let x = the number of cookies each neighbor will receive

Let y = the total number of baked cookies

⇒ y = 12x

⇒ 125 = 12x

⇒ x = 125 ÷ 12

⇒ x = 10.42

Therefore, each neighbor will receive 10 cookies (we have to round it down, as if each neighbor received 11 cookies, we would need 132 in total).

Let x = the number of cheesecakes purchased

Let y = the amount of cheescake each guest will receive

⇒ y = 125x

⇒ 12 = 125x

⇒ x = 12 ÷ 125

⇒ x = 12/125 = 0.096 = 9.6%

Therefore, each guest will receive 12/125ths of a cheesecake (or 9.6%).

6 0

2 years ago