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2 years ago

I can not figure out how to do this question and I need help please

Mathematics

1 answer:

spayn [35]2 years ago

6 0

$\bf \textit{quadratic formula}\\\\

\begin{array}{lcclll}
1x^2&-6x&+12&=0\\
\uparrow &\uparrow &\uparrow \\
a&b&c
\end{array}
\qquad \qquad 
x= \cfrac{ - {{ b}} \pm \sqrt { {{ b}}^2 -4{{ a}}{{ c}}}}{2{{ a}}}$

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Answer:

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3 years ago

What is 2/10 times 20 as a fraction?

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2 years ago

Help me please !!!!!

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2 years ago

Vector u has a magnitude of 5sq rt 3 feet and is directed due west. Vector v has a magnitude

Vikki [24]

The magnitude and the direction angle for the resultant vector are 10N and 120degrees respectively

The formula for calculating the resultant vector is expressed as:

$R^2 = u^2 + v^2\\R =\sqrt{u^2+v^2}$

Given the following distances:

u = -5√3 feet (west direction)

v = 5 feet

Substitute the given values into the formula above;

$R=\sqrt{(5\sqrt{3} )^2+5^2} \\R=\sqrt{(25 \times 3)+25}\\R=\sqrt{75 + 25}\\R=\sqrt{100}\\R = 10 feet$

The magnitude of the resultant vector is 10feet

Get the direction:

$\theta = tan^{-1}\frac{y}{x} \\\theta = tan^{-1}\frac{5}{-5\sqrt{3} }\\\theta =tan^{-1}\frac{-1}{\sqrt{3} }\\\theta = -60^0$

Since tan is negative in the second quadrant, hence;

$\theta = 180 - 60\\\theta = 120^0$

Hence the magnitude and the direction angle for the resultant vector are 10N and 120degrees respectively.

Learn more here: brainly.com/question/5177683

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2 years ago