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Elis [28]
3 years ago
12

The section inside the border is x inches and (12-x) inches wide. How do I write a function to represent the area a of the secti

on inside the border
Mathematics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

The function to represent the area a of the section inside the border is A=12x-x^2

Step-by-step explanation:

Given section inside the border is x inches and (12-x) inches.

We need to write the function to represent the area of the section inside the border.

We can see the length of border is x and width of the border is (12-x).

So, the area of the rectangle will be length\times width

The area (A) will be

A=x(12-x)\\\\A=12x-x^2

So, the function to represent the area a of the section inside the border is A=12x-x^2

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3 years ago
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From a window 20 feet above the ground, the angle of elevation to the top of a building across
Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation:  Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across  the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},

and from the right-angled triangle WAB, we have'

\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.

Therefore, AH = AB + BH = h + GB = 361.85+20 = 381.85 feet.

Thus, the height of the building across the street is 381.85 feet.

8 0
3 years ago
Write the new equation if the graph of f(x)= 4x^2 + 3 is<br> translated up 4 units.
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Its impossible to solve this, but i got -47
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What are the graphs of y=sinx and y=cosx in the interval from -2 pi to 2 pi?
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I used Excel to create the tables and the graphs.

I attached both the tables and the graphs.


You should replace the numbers of the x-axis (in both graphs) by the numbers as a fraction of pi. Those numbers are also included in the table, so you should not have problems with that.


Open and see tha file attached with the answer to your question..
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5 0
3 years ago
Evaluate the expression Arcsec(sqrt 2). Do not use your calculator to answer this question. Do not express answer in decimals.
Igoryamba
Pi/4 radians  
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 sqrt(2) = 1/x
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 Let's multiply both numerator and denominator by sqrt(2), so
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3 years ago
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