The area of a rectangle is 47.5 square centimeters. The

2 answers:

7 0

SHORT ANSWER: 84cm

Let width = w

Let length = w + 5

The area of a rectangle is found by multiplying its width by length and the problem states that the area is 84 cm2.

(Width)(Length) = 84

(w)(w + 5) = 84

w2 + 5w = 84

w2 + 5w - 84 = 0

(w + 12)(w -7) = 0

w + 12 = 0

w = -12

The width of a rectangle can't be NEGATIVE!!! Forget about that!

w - 7 = 0

w = 7

Now go back to naming the variables and replace the w with 7.

Length = w + 5 = 7 + 5 = 12

Width is 7 cm. Length is 12 cm.

Check it. Is 7 x 12 = 84? Yes.

pychu [463]3 years ago

6 0

Answer:

Step-by-step explanation:

You have to divide 47.5 and 5

Hope this helps lol

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jenyasd209 [6]

<h3>Answers:</h3><h3>There are four solutions and they are</h3>

$\theta = 0$
$\theta = \pi$
$\theta = \frac{7\pi}{6}$ ... this says "7pi over 6"
$\theta = \frac{11\pi}{6}$ ... this says "11pi over 6"

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Work Shown:

$\sin(\theta)+1 = \cos(2\theta)\\\\\sin(\theta)+1 = 1-2\sin^2(\theta)\\\\\sin(\theta)+1-1+2\sin^2(\theta)=0\\\\2\sin^2(\theta)+\sin(\theta)=0\\\\\sin(\theta)(2\sin(\theta)+1)=0\\\\\sin(\theta)=0 \ \text{ or } \ 2\sin(\theta)+1=0\\\\$

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Solving $\sin(\theta)=0$ leads to $\theta=n\pi$ for any integer n.

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Solving $2\sin(\theta)+1=0$ leads to...

$2\sin(\theta)+1=0\\\\2\sin(\theta)=-1\\\\\sin(\theta)=-\frac{1}{2}\\\\\theta=\arcsin\left(-\frac{1}{2}\right) \ \text{ or } \ \theta=\pi-\arcsin\left(-\frac{1}{2}\right)\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\pi-\frac{11}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{-5}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{7}{6}\pi+2\pi n\\\\$

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The general solution set is

$\{\theta=n*\pi, \theta=\frac{11}{6}\pi+2\pi n, \theta=\frac{7}{6}\pi+2\pi n\}$

Again, n is any integer.

Let's look at a table of values where we plug in various integers for n. See the attached image below. Note the stuff in the highlighted yellow cells represents expressions that are between 0 and 2pi = 6.28

Therefore the four solutions are $\{0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\}$ after we plug the proper values of n into the expressions, to have things match what the table shows.