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11111nata11111 [884]

3 years ago

14

Describe the vector below shown below as an ordered pair. X-value? Y-value?

2 answers:

Cloud [144]3 years ago

6 0

$\bf \begin{cases}
x=rcos(\theta )\\
y=rsin(\theta )\\
------\\
r=22\\
\theta =39
\end{cases}\implies (\stackrel{x}{22\cdot cos(39^o)}~,~\stackrel{y}{22\cdot sin(39^o)})
\\\\\\
(x,y)\approx (17.09721115~~,~~13.84504860)$

rosijanka [135]3 years ago

3 0

The red vector has a magnitude of 22 and makes an angle of 39 degrees with the horizonal x-axis.

Converting to rectangular form: x = 22 cos 39 deg and y = 22 sin 39 deg

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Using the quadratic formula to solve 4x2 – 3x + 9 = 2x + 1, what are the values of x?

yaroslaw [1]

Answer:

The value of x is:

$x=\dfrac{5\pm \sqrt{103}i}{8}$

Step-by-step explanation:

we have to use the quadratic formula to solve for x.

The equation is given as:

$4x^2-3x+9=2x+1$

which could also be written as:

$4x^2-3x+9-2x-1=0\\\\4x^2-3x-2x+9-1=0\\\\\\4x^2-5x+8=0$

The quadratic formula for the quadratic equation of the type:

$ax^2+bx+c=0$ is given as:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here we have:

a=4, b=-5 and c=8.

Hence, by the quadratic formula we have:

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4\times 8\times 4}}{2\times 4}\\\\x=\dfrac{5\pm \sqrt{25-128}}{8}\\\\\\x=\dfrac{5\pm \sqrt{103}i}{8}$

Hence, the value of x is:

$x=\dfrac{5\pm \sqrt{103}i}{8}$

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