$\bullet \ cos(28^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{A_{x}}{44} \\ \\ \therefore A_{x}=44cos(28^{\circ})=38.85m \\ \\ \\ \bullet \ sin(28^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{A_{y}}{44} \\ \\ \therefore A_{y}=44sin(28^{\circ})=20.65m$

$\bullet \ cos(56^{\circ})=\frac{Adjacent}{Hypotenuse}=\frac{-B_{x}}{26.5} \\ \\ \therefore B_{x}=-26.5cos(56^{\circ})=-14.81m \\ \\ \\ \bullet \ sin(56^{\circ})=\frac{Opposite}{Hypotenuse}=\frac{B_{y}}{26.5} \\ \\ \therefore B_{y}=26.5sin(56^{\circ})=21.97m$

Therefore:

$\vec{A}=(38.85, 20.65)m \\ \\ \vec{B}=(-14.81, 21.97)m$

So:

$\vec{B}-\vec{A}=(-14.81, 21.97)-(38.85, 20.65)=(-53.66,1.32)$

Finally, the magnitude is:

$\boxed{\left| \vec{B}-\vec{A}\right|=\sqrt{(-53.66)^2+(1.32)^2}=53.67m}$

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2 years ago

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If f(x) = 7 + 4x and g(x)= 7x, what is the value of (f/g)(5)

andrew-mc [135]

The value of $(\frac{f}{g})(5)$ is $\frac{27}{35}$

Explanation:

Given that $f(x)=7+4x$ and $g(x)=7x$

<u>The value of </u> $(\frac{f}{g})(5)$ <u>:</u>

The value of $(\frac{f}{g})(5)$ can be determined by substituting x = 5 in the functions f(x) and g(x) and then dividing the terms, we get, the value of $(\frac{f}{g})(5)$

Thus, we have,

$(\frac{f}{g})(5)=\frac{f(5)}{g(5)}$