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seraphim [82]
4 years ago
4

John has a blueprint for a square-shaped office. He decides to change the layout and so draws a new blueprint such that the widt

h of his office increased by 15 feet. The area of his office according to the new blueprint is given by the expression below, where x is the side length, in feet, of the original square-shaped office.
x^2+15x

Which statement best describes the term 15x?

the area of the office according to the old blueprint
the width of the office according to the new blueprint
the area added to the office according to the new blueprint
the area of the office according to the new blueprint

Mathematics
2 answers:
rewona [7]4 years ago
8 0
See the figure attached

we know that

the <span>old blueprint is the square ABCD
Area=x*x=x</span>²----------A1=x²

<span>the new blueprint is the rectangle AEFB
Area=(x+15)*x----------> A2=x</span>²+15*x

so

A2=A1+15*x

therefore

the answer is 
<span>the term 15x is </span>the area added to the office according to the new blueprint


kobusy [5.1K]4 years ago
4 0

Answer:

The area added to the office according to the new blueprint

Step-by-step explanation:

you're welcome

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Answer:

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b).

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c).

(fog)( - 2) = 3( - 2) + 5 \\  =  - 6 + 5 \\  =  - 1

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(gof)( - 2) = 3( - 2) + 13 \\  =  - 6 + 13 \\  = 7

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Y=4


Don’t judge if I’m wrong

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Answer:

  • cos(x) = a/c
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Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to help you remember the relationships between triangle sides and trig functions. These abbreviations tell you that ...

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Oksanka [162]

Answer:

\sin L = \frac{3}{5}

\tan N = \frac{4}{3}

\cos L = \frac{4}{5}

\sin N = \frac{4}{5}

Step-by-step explanation:

Given

The above triangle

First, we calculate the length LM using Pythagoras theorem.

LN^2 = LM^2 + MN^2

10^2 = LM^2 + 6^2

100 = LM^2 + 36

Collect like terms

LM^2 = 100 - 36

LM^2 = 64

Take positive square root

LM=8

Solving (a): Sin L

\sin L = \frac{Opposite}{Hypotenuse}

\sin L = \frac{MN}{LN}

\sin L = \frac{6}{10}

Simplify

\sin L = \frac{3}{5}

Solving (b): tan N

\tan N = \frac{Opposite}{Adjacent}

\tan N = \frac{LM}{MN}

\tan N = \frac{8}{6}

Simplify

\tan N = \frac{4}{3}

Solving (c): cos L

This calculated as:

\cos L = \frac{Adjacent}{Hypotenuse}

\cos L = \frac{LM}{LN}

\cos L = \frac{8}{10}

Simplify

\cos L = \frac{4}{5}

Solving (d): sin N

This is calculated using:

If a + b = 90

Then: \sin a = \cos b

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\sin N = \cos L

\sin N = \frac{4}{5}

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