All categories

3 years ago

I spent all my points into this so give it your best shot

Mathematics

2 answers:

ioda3 years ago

5 0

Answer: hi

Step-by-step explanation:

pshichka [43]3 years ago

5 0

Answer:

yea he right trust me deshecho detrás amigo latina español

You might be interested in

Which region labeled in the graph below would represent the solution (the final shaded area) to the system of linear inequalitie

Elena L [17]

Answer:

Region D.

Step-by-step explanation:

Here we have two inequalities:

y ≤ 1/2x − 3

y < −2/3x + 1

First, we can see that the first inequality has a positive slope and the symbol (≤) so the values of the line itself are solutions, this line is the solid line in the graph.

And we have that:

y ≤ 1/2x − 3

y must be smaller or equal than the solid line, so here we look at the regions below the solid line, which are region D and region C.

Now let's look at the other one:

y < −2/3x + 1

y = (-2/3)*x + 1

is the dashed line in the graph.

And we have:

y < −2/3x + 1

So y is smaller than the values of the line, so we need to look at the region that is below de dashed line.

The regions below the dashed line are region A and region D.

The solution for the system:

y ≤ 1/2x − 3

y < −2/3x + 1

Is the region that is a solution for both inequalities, we can see that the only region that is a solution for both of them is region D.

Then the correct option is region D.

5 0

3 years ago

MATH HELP!!! 100PTS!!!

Lostsunrise [7]

Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

Given functions:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the quadratic formula and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$