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2 years ago

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The ratio of the height to the width of a packaging label is 5 to 19. If the height of the label is 2 inches, what is its width?

HINT: Set up and solve as a proportion and find x.

2 answers:

Svetradugi [14.3K]2 years ago

6 0

Answer:

7.6 inches

Step-by-step explanation:

The ratio of the height to the width of a packaging label is 5 to 19, that is,

h/w = 5/19 [1]

If the height of the label is 2 inches, and its width is unknown (x), that is,

h/w = 2 in/x [2]

We can set up a proportion with [1] and [2],

5/19 = 2 in/x

We use the cross product property of proportions.

5 × x = 19 × 2 in

x = 19 × 2 in/ 5 = 7.6 in

The width is 7.6 inches.

murzikaleks [220]2 years ago

3 0

Height/width=5/19
5/19=2/x
(5)(x)=(19)(2)
5x=38
5/5x=38/5
x=38/5in or 7.6in

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F(x)=3x 2 +9f, left parenthesis, x, right parenthesis, equals, 3, x, squared, plus, 9 and g(x)=\dfrac{1}{3}x^2-9g(x)= 3 1 x 2

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Answer:

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

$g(f(x)) = 3x^4 + 18x^2 + 18$

<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

$f(x) = 3x^2 + 9$

$g(x) = \dfrac{1}{3}x^2 - 9$

Required

Determine f(g(x))

Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

$f(x) = 3x^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9$

$f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9$

Expand Brackets

$f(g(x)) = (x^2 - 27)(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = x^2(\frac{1}{3}x^2 - 9) - 27(\frac{1}{3}x^2 - 9) + 9$

$f(g(x)) = \frac{1}{3}x^4 - 9x^2 - 9x^2 + 243 + 9$

$f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252$

Calculating g(f(x))

$g(x) = \dfrac{1}{3}x^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9$

$g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9$

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Expand Brackets

$g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9$

$g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9$

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Checking for inverse functions

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Represent f(x) with y

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Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

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$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

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$y^2 = \frac{x}{3} - 3$

Take square root of both sides

$\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}$

$y = \sqrt{\frac{x}{3} - 3}$

Represent y with g(x)

$g(x) = \sqrt{\frac{x}{3} - 3}$

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