Yes, i<span>n mathematics, a </span>rational number<span> is any </span>number<span>that can be expressed as the quotient or fraction p/q of two integers, a numerator p and a non-zero denominator q.</span>
Answer:
14+=+2L+%2B+2%28L-3%29 Simplify and solve for L
14+=+2L+%2B+2L+-+6 Combine like-terms.
14+=+4L-6 Add 6 to both sides.
20+=+4L Divide both sides by 4.
5+=+L The length is 5 meters.
W+=+L-3
W+=+5-3
W+=+2meters.
The width is 2 meters.
P+=+2L%2B2W
P+=+2%285%29%2B2%282%29
P+=+10%2B4
P+=+14meters.
Step-by-step explanation:
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)

<u><em>Answer:</em></u>
The initial value is 3
It represents the distance between the bird and its nest before any elapsed time which is 3 miles.
<u><em>Step-by-step explanation:</em></u>
The graph is shown in the attached image.
<u>The general form of a straight line is:</u>
y = mx + c
where:
m is the slope and c is the y-intercept.
The y-intercept is also known as the initial value. It is the value of y at x=0.
From the givens, we can note that at x = 0, the y-value is 3. This means that the y-intercept is 3.
Now, in the problem, x represents the time in minutes while y represents the distance of the bird from the nest in miles.
At x = 0, meaning before any time elapsed, the distance of the bird from its nest was 3 miles.
Hope this helps :)