Ask question

Ask question

All categories

2 years ago

5

Determine the slope-intercept form of the equation of the line parallel to y = -4/3 x + 11 that passes through the point (–6, 2)

. y = x +

1 answer:

rewona [7]2 years ago

6 0

Answer: -4/3x - 6

Step-by-step explanation:

First, let's find the slope of the line

y=- -4/3x+11

As the equation is already in slope-intercept form y=mx+c ,

Slope = -4/3

Let a point (x,y) be on the new line.

By finding the slope again,

y−2/x+6= -4/3

y−2= -4/3(x+6)

y−2= -4/3x−8

y = -4/3x - 6

You might be interested in

Elimination_4x_2y=14<br> _10x+7y=_24

Gnesinka [82]

Answer:

The answer is $x=-\frac{50}{8}$ and $y=\frac{11}{2}.$

Step-by-step explanation:

Given:

-4x-2y=14

-10x+7y=-24

Now, to solve it by elimination:

$-4x-2y=14$ ......(1)

$-10x+7y=-24$ ......(2)

So, we multiply the equation (1) by 7 we get:

$-28x-14y=98$

And, we multiply the equation (2) by 2 we get:

$-20x+14y=-48$

Now, adding both the new equations:

$-28x-14y+(-20x+14y)=98+(-48)$

$-28x-14y-20x+14y=98-48$

$-28x-20x-14y+14y=50$

$-8x=50$

<em>Dividing both the sides by -8 we get:</em>

$x=-\frac{50}{8}$

Now, putting the value of $x$ in equation (1):

$-4x-2y=14$

$-4(-\frac{50}{8})-2y=14$

$\frac{200}{8} -2y=14$

$25-2y=14$

<em>Subtracting both sides by 25 we get:</em>

$-2y=-11$

<em>Dividing both sides by -2 we get:</em>

$y=\frac{11}{2}$

Therefore, the answer is $x=-\frac{50}{8}$ and $y=\frac{11}{2}.$

5 0

3 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

3 years ago

What is the sum of 4.34

nirvana33 [79]

What is the rest of the expression?

3 0

2 years ago

Add. Write fractions in simplest form.<br> 4-12 EVENS

Virty [35]

4. 1/3 6.-0.9 or -9/10 8. 2 1/3 10. -3.45 12.4.54 or 4 27/50

7 0

2 years ago

Y+9÷2=y-3÷4-(y/3)<br>pls help

Ann [662]

Answer:

y = - $\frac{63}{4}$ or - 15.75

Step-by-step explanation:

Order of operations.

y + $\frac{9}{2}$ = y - $\frac{3}{4}$ - $\frac{y}{3}$ Subtract y from each side.

y - y + $\frac{9}{2}$ = y - y - $\frac{3}{4}$ - $\frac{y}{3}$

$\frac{9}{2}$ = - $\frac{3}{4}$ - $\frac{y}{3}$ Add $\frac{3}{4}$ to each side

$\frac{9}{2}$ + $\frac{3}{4}$ = - $\frac{3}{4}$ + $\frac{3}{4}$ - $\frac{y}{3}$

$\frac{9}{2}$ + $\frac{3}{4}$ = - $\frac{y}{3}$ Find the common denominator for $\frac{9}{2}$ and $\frac{3}{4}$ , which is 4

$\frac{9}{2}$ * $\frac{2}{2}$ + $\frac{3}{4}$ = - $\frac{y}{3}$

$\frac{18}{4}$ + $\frac{3}{4}$ = - $\frac{y}{3}$

$\frac{21}{4}$ = - $\frac{y}{3}$ Multiply each side by 3

$\frac{21}{4}$ * 3 = - $\frac{y}{3}$ * 3

$\frac{63}{4}$ = - y Divide each side by -1

y = - $\frac{63}{4}$ = - 15 $\frac{3}{4}$

y = - 15.75

3 0

3 years ago

Read 2 more answers