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2 years ago

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A rectangle has a height of w^2+3w+9w 2 +3w+9w, squared, plus, 3, w, plus, 9 and a width of w^2+2w 2 +2w, squared, plus, 2. Expr

ess the area of the entire rectangle.

1 answer:

Keith_Richards [23]2 years ago

4 0

Answer:

w⁴+5w³+17w²+24w +18

Step-by-step explanation:

Area of a rectangle = length * width

Given

Length = w²+3w+9

width = w²+2w+2

Area of the rectangle = (w²+3w+9)(w²+2w+2)

Area of the rectangle = w²(w²) + 2w(w²) + 2w² + 3w(w²) + 3w(2w) + 2(3w)+9w²+9(2w)+9(2)

Area of the rectangle = w^4+2w^3+2w²+3w^3+6w^2+6w+9w²+18w+18

Collect the like terms;

Area of the rectangle = w^4+2w^3+3w^3+2w²+6w²+9w²+6w+18w+18

Area of the rectangle = w⁴+5w³+17w²+24w +18

Hence the area of the entire triangle is w⁴+5w³+17w²+24w +18

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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:

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

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<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))

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

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

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

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

Expand Brackets

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

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

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

Checking for inverse functions

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

Represent f(x) with y

$y = 3x^2 + 9$

Swap positions of x and y

$x = 3y^2 + 9$

Subtract 9 from both sides

$x - 9 = 3y^2 + 9 - 9$

$x - 9 = 3y^2$

$3y^2 = x - 9$

Divide through by 3

$\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}$

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

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

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Answer:

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Step-by-step explanation:

Division in a ratio of 1:2:3

1 + 2 + 3 = 6

So

The first box will have 1/6 of the marbles.

The second box will have 2/6 = 1/3 of the marbles

The third box will have 3/6 = 1/2 of the marbles.

First box:

One sixth, so:

(1/6)*540 = 540/6 = 90

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One third, so:

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Third box:

One half, so:

(1/2)*540 = 540/2 = 270

The first box will have 90 marbles, the second 180 and the third 270.

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