well, this is just a matter of simple unit conversion, so let's recall that one revolution on a circle is just one-go-around, radians wise that'll be 2π, and we also know that 1 minute has 60 seconds, let's use those values for our product.
![\cfrac{300~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{2\pi ~rad}{~~\begin{matrix} r \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }\cdot \cfrac{~~\begin{matrix} min \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{60secs}\implies \cfrac{(300)(2\pi )rad}{60secs}\implies 10\pi ~\frac{rad}{secs}\approx 31.42~\frac{rad}{secs}](https://tex.z-dn.net/?f=%5Ccfrac%7B300~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B2%5Cpi%20~rad%7D%7B~~%5Cbegin%7Bmatrix%7D%20r%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%5Ccdot%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20min%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%20%7D%7B60secs%7D%5Cimplies%20%5Ccfrac%7B%28300%29%282%5Cpi%20%29rad%7D%7B60secs%7D%5Cimplies%2010%5Cpi%20~%5Cfrac%7Brad%7D%7Bsecs%7D%5Capprox%2031.42~%5Cfrac%7Brad%7D%7Bsecs%7D)
9n + 6 = 456
9n = 456 - 6
9n = 450
n = 450/9
n = 50
I believe this is the answer, though i'm not sure.
Answer:
ok dude
Step-by-step explanation:
Wʜᴇʀᴇ ɪs ǫᴜᴇsᴛɪᴏɴ?
We know that
<span>In a right triangle
if A and B are complementary angles
cos A=sin B
therefore
if cos A=5/13
then sin B=5/13
the answer is
sin B=5/13</span>
Answer:
DL/dt = 17 mph
Step-by-step explanation:
The ships and the port shape a right triangle
Ship going west ( x-direction) is traveling at 15 mph
Ship going north (y-direction) is traveling at 10 mph
The distance L between ships is:
L² = x² + y²
Tacking derivatives on both sides of the equation with respect to time
we get
2*L*DL/dt = 2*x*Dx/dt + 2*y*Dy/dt (1)
In that equation we know:
Dx/dt = 15 mph
Dy/dt = 10 mph
At the moment ship traveling from the east is at 3 miles from the port
then x = 3 m and the other ship is at 4 miles north
then by Pythagoras theorem
L = √ 3² + 4² L = 5
By substitution in equation 1
2*5*DL/dt = 2*3*15 + 2*4*10
10* DL/dt = 90 + 80
DL/dt = 170 / 10
DL/dt = 17 mph
