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3 years ago

How are <1 and <4 related , please help I would appreciate it

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

4 0

They are both vertical angles.

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In which of the following quadrilaterals are consecutive and opposite angles always congruent?

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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:

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

Therefore, the solutions are:

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

Learn more about quadratic equations here:

brainly.com/question/27750885

brainly.com/question/27739892

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