What is √18x^6 written in simplest for? assume all variables are positive.

Which point would be a solution to the system of linear inequalities shown below? y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1 y&g

Sauron [17]

Answer:

$x < \frac{20}{17}$ and $y \ge \frac{-5}{17}$

Step-by-step explanation:

Given

$y>4 x-5\hspace{50px}y\ge\frac{3}{5} x-1$

Required

Find x and y

In the second equation. Assume that:

$y = \frac{3}{5}x - 1\\$

Substitute $y = \frac{3}{5}x - 1$ in the first equation

$y > 4x - 5$

$\frac{3}{5}x - 1 > 4x - 5$

Collect like terms

$\frac{3}{5}x - 4x > - 5 + 1$

$\frac{3}{5}x - 4x > -4$

Multiply through by 5

$3x - 20x > -20$

$-17x > -20$

Solve for x

$x < \frac{20}{17}$

Substitute this value of x in $y \ge \frac{3}{5}x - 1$

$y \ge \frac{3}{5}*\frac{20}{17} - 1$

$y \ge \frac{3}{1}*\frac{4}{17} - 1$

$y \ge \frac{12}{17} - 1$

$y \ge \frac{12- 17}{17}$

$y \ge \frac{-5}{17}$

3 0

3 years ago

What is the equation of the line that passes through the points (-4,1) and (4-1)

Nitella [24]

Answer:

y=-1/4x

Step-by-step explanation:

METHOD 1: The basic slope-intercept equation is <em>y=ax+b</em>. To achieve the slope with two points, this is the equation:

$m = \frac{y2-y1}{x2-x1}$

All you have to do is insert points (-4, 1) and (4, -1) into their corresponding locations.

$m = \frac{(-1)-1}{4-(-4)}$

With this equation, you will be able to find the slope. Then you can simply graph it to connect the two dots.

METHOD 2: Simply connect the two dots with a straight line (use a ruler or something with a straight edge.) Then count how many spaces up/down and left/right. If the line goes from top to bottom from left to right, it is a negative slope (hence, the -1/4.)

7 0

3 years ago

30 POINTS,NEED HELP ASAP !!!

zaharov [31]

Answer: OPTION A.

Step-by-step explanation:

You can observe that in the figure CDEF the vertices are:

$C(-2,-1),\ D(-2,0),\ E(2,2)\ and\ F(2,1)$

And in the figure C'D'E'F' the vertices are:

$C'(-8,-4),\ D'(-8,0),\ E'(8,8)\ and\ F'(8,4)$

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:

For C'(-8,-4) and C(-2,-1):

$\frac{-8}{-2}=4\\\\\frac{-4}{-1}=4$

Let's choose another vertex. For E'(8,8) and E(2,2):

$\frac{8}{2}=4\\\\\frac{8}{2}=4$

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.

Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:

$(x, y)$ → $(4x, 4y)$