$^{(1)}\ \ 4x^5-12x^4+6x=5x^3-2x\ \ ^{(2)}\\\\ \left\{\begin{array}{ccc}y=4x^5-12x^4+6x\ \ \ ^{(1)}\\y=5x^3-2x\ \ \ ^{(2)}\end{array}\right$

Answer: system nr 4.

7 0

2 years ago

Write your answer with no space and any fraction with “/“

nirvana33 [79]

Answer:

y+3=2(x-1)

Step-by-step explanation:

Point-slope form is $y-y_1 = m (x-x_1)$ . In order to write a linear equation using this form, the $y_1$ , $x_1$ and $m$ must be substituted by real numbers.

The $m$ represents the slope of the equation. We know that the answer has to be parallel to the line y=2x-1. y=2x-1 is already in slope-intercept form, or $y=mx+b$ , in which $m$ represents the slope. We can see that the $2$ is in place of that $m$ , therefore 2 is the slope of y=2x-1. Lines that are parallel have the same slope, so we know that the slope of the new equation must be 2 as well.

The $x_1$ and $y_1$ represent the x and y values of one point that the line passes through. We know that the new equation must pass through the point (1, -3), so we'll substitute 1 in place of the $x_1$ and -3 in place of the $y_1$ .

Substituting for those three values will give you an equation like this, therefore it is the answer:

$y-(-3)= 2(x-(1))$

$y+3= 2(x-1)$

4 0

2 years ago