Answer:
B) 54π in²
Step-by-step explanation:
area of full circle = πr² = 81π in²
area of 4π/3 segment = (81)(4π/3)/2π = 54π in²
Answer:
Step-by-step explanation:
The area of a rectangle can be found with the formula
, where
is the length of the rectangle and
is the width.
From the given problem statement, we know that
and
, so we can fill in those values in the formula anbd solve for
to get the width:


Divide both sides by
to isolate
by itself:


Answer:
Step-by-step explanation:
y-y1 = m(x-x1)
y-4 = 2(x-3)
y-4=2x-6
y= 2x -6+4
y=2x-2
Answer:

Step-by-step explanation:
Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have
Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising
<em>P(0) = 0 and P(50) = 1,500,000
</em>
We have and ordinary differential equation of first order that we can write
The <em>integrating factor </em>is
Multiplying both sides of the equation by the integrating factor
Hence
Integrating both sides
But P(0) = 0, so C = -3,000,000
and P(50) = 1,500,000
so
And the equation that models the number of people (in millions) who become aware of the product by time t is
let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
