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monitta
2 years ago
5

Simon has a certain length of fencing to enclose a rectangular area. The function AAA models the rectangle's area (in square met

ers) as a function of its width (in meters).
Which of these statements are true?

Choose all answers that apply:

Choose all answers that apply:


(Choice A)

A

Greater width always relates to smaller area.


(Choice B)

B

Greater width relates to smaller area as long as the width is less than 10\text{ m}10 m10, start text, space, m, end text.


(Choice C)

C

When there is no width, the area is 20\text{ m}^220 m

2

20, start text, space, m, end text, squared.


(Choice D)

D

When there is no width, the area is 0\text{ m}^20 m

2

0, start text, space, m, end text, squared.
Mathematics
1 answer:
pav-90 [236]2 years ago
4 0

Answer:

When there is no width, the area is 0m²

Step-by-step explanation:

Expressing the area of a rectangle as a function of its width:

Area of a rectangle = Length * width

Since the area of rectangle is related to the width by the multiplication operatkr. We can conclude from the options given that ; if the value of width = 0 ; the the area of rectangular plot = 0

When x :

Area = Length * width

Area = Length * 0

Area = 0

Area = Length * 0

Aea = 0

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Find the exact value of tan(165°) using a difference of two angles
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Answer:  -2+\sqrt{3}

=========================================================

Work Shown:

Apply the following trig identity

\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\

Now let's rationalize the denominator

\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\

----------------------

As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as \sqrt{3}-2 but it's the same as above.

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