A ship leaves port at noon and travels at a bearing of 214°. The ship’s average rate of speed is 15 miles per hour.
Describe the location of the ship at 2:30 p. m. by writing a vector in component form.

The component form of the vector representing the ship at 2:30 p. m. is approximately ⟨ , ⟩.

(-31.09 mi, -20.97 mi)

Step-by-step explanation:

If the noon is our initial time, then we have that 2:30 pm is t = 2.5 hours.

Now, the position (x, y) of the ship can be writen as:

(velocity*time*cos(angle), velocity*time*sin(angle))

or:

(15mph*t¨*cos(214°), 15mph*t*sin(214°))

then the position at t = 2.5 hours is:

(15mph*2.5h*cos(214°), 15mph*2.5h*sin(214°))

(-31.09 mi, -20.97 mi)

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204