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mel-nik [20]
2 years ago
11

A certain television is advertised as a 55-inch TV (the diagonal length). If the width of the TV is 42 inches, how many inches t

all is the TV? Round your answer to the nearest tenth.
Mathematics
1 answer:
Mumz [18]2 years ago
7 0

Answer:

69.20in

Step-by-step explanation:

Given data

diagonal of the TV d= 55in

Widht of the TV w=  42 in

Hight h= ???

Applying the Pythagoras theorem

d^2= w^2+ h^2

substitute

55^2= 42^2+ h^2

3025= 1764+ h^2

3025+1764= h^2

4789= h^2

h=√4789

h= 69.20

Hence the Heigth is 69.20in

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Answer:

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Step-by-step explanation:

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3 years ago
For the equation y=7x+3 the value of X changes from 1 to 2
WITCHER [35]

When x=1, 7x+3 = 10.

When x=2, 7x+3 = 17.

When the value of 'x' changes from '1' to '2', the value of 7x+3 changes
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That's what the ' 7x ' in the equation means, and that's what we mean
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3 years ago
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GrogVix [38]

Answer: (-6, -16)

Step-by-step explanation:

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6x-3y=12

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3 years ago
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Hello,
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Rewrite 2tan3x in terms of tanx
blsea [12.9K]
2\tan3x=2\tan(2x+x)=\dfrac{2\tan2x+2\tan x}{1-\tan2x\tan x}

Use the same identity to expand \tan2x.

\tan2x=\tan(x+x)=\dfrac{\tan x+\tan x}{1-\tan x\tan x}=\dfrac{2\tan x}{1-\tan^2x}

\implies2\tan3x=\dfrac{2\dfrac{2\tan x}{1-\tan^2x}+2\tan x}{1-\dfrac{2\tan x}{1-\tan^2x}\tan x}
2\tan3x=\dfrac{2\tan x\left(\dfrac2{1-\tan^2x}+1\right)}{1-\dfrac{2\tan^2x}{1-\tan^2x}}
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