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3 years ago

write the equation of the line perpendicular to 3x + y = 4, containing the point (-6, -1). Write in slope intercept form.

Mathematics

2 answers:

Lesechka [4]3 years ago

7 0

Y=3x+y is in slope intercept form

Advocard [28]3 years ago

4 0

First thing you need to do is find the slope. The slope must be opposite in sign and flipped.

You might be interested in

Line p has a slope of 6/7. Line q is perpendicular to p . What is the slope of line q

Serjik [45]

Answer:

The slope of line q is -7/6

Step-by-step explanation:

To find the slope of a perpendicular line, you need to find the opposite reciprocal. So, flip the fraction (7/6) and then take the opposite, (-7/6).

Hope this helps!

7 0

3 years ago

Helpppo:)))))) pleaseeee

velikii [3]

Answer:

Step-by-step explanation:

Step 1: Define

f(x) = 3x² - 5x - 4

g(x) = -4x - 12

Step 2: Find f(g(x))

f(g(x)) = 3(-4x - 12)² - 5(-4x - 12) - 4

f(g(x)) = 3(16x² + 96x + 144) + 20x + 60 + 4

f(g(x)) = 48x² + 288x + 432 + 20x + 64

f(g(x)) = 48x² + 308x + 496

Step 3: Find f(g(-4))

f(g(-4)) = 48(-4)² + 308(-4) + 496

f(g(-4)) = 48(16) - 1232 + 496

f(g(-4)) = 768 - 736

f(g(-4)) = 32

3 0

3 years ago

How do you know if you are moving the object positive or negative on the x-axis?

Nitella [24]

Answer:

Well I think positive is to the right and negative is to the left

3 0

3 years ago

What property says that -12+12+0

natali 33 [55]

Answer:

Looks like <u>inverse property of addition </u>to me

4 0

3 years ago

For the system of equations that follows, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in

Rainbow [258]

Answer:

$\begin{Vmatrix}1 & 0 & 0& 46/41 &180/41\\0 & 1 & 0&1/41 &-5/41\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

Step-by-step explanation:

From the question we are told that

System of equations given as

x₁ + 3x₂ + x₃ + x₄ = 3;

2x₁ - 2x₂ + x₃ + 2x₄ = 8;

x₁ - 5x₂ + x₄ = 5

Matrix form

$\begin{Vmatrix}1 & 3 & 1&1&3\\2 & -2 & 1&2&8\\0&1 & -5 & 1& 5\end{Vmatrix}$ $\begin{Vmatrix}x_1\\x_2\\x_3\\x_4\end{Vmatrix}$

Generally the echelon reduction is mathematically applied as

$\begin{Vmatrix}1 & 3 & 1&1&3\\2 & -2 & 1&2&8\\0&1 & -5 & 1& 5\end{Vmatrix}$

Add -2 times the 1st row to the 2nd row

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & -8 & -1&0&2\\0&1 & -5 & 1& 5\end{Vmatrix}$

Multiply the 2nd row by -1/8

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & 1 & 1/8&0&-1/4\\0&1 & -5 & 1& 5\end{Vmatrix}$

Add -1 times the 2nd row to the 3rd row

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & 1 & 1/8&0&-1/4\\0&0 & -41/8 & 1& 21/4\end{Vmatrix}$

Multiply the 3rd row by -8/41

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & 1 & 1/8&0&-1/4\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

Add -1/8 times the 3rd row to the 2nd row

Add -1 times the 3rd row to the 1st row

$\begin{Vmatrix}1 & 3 & 0&49/41&165/41\\0 & 1 & 0&1/41 &-5/41\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

Add -3 times the 2nd row to the 1st row

$\begin{Vmatrix}1 & 0 & 0& 46/41 &180/41\\0 & 1 & 0&1/41 &-5/41\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

3 0

3 years ago