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3 years ago

PLEASE HELP KZKKSOvnfkdjd for sloeidkcncnsk get callallapwd

Mathematics

1 answer:

astraxan [27]3 years ago

7 0

S = -3...

get rid of your fraction by multiplying everything by 12...

now minus each side by 60

that leaves you -3 because the right side was 57.

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2 years ago

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Svet_ta [14]

Answer:

$A = \int\limits^3__-3}{9}-{x^{2}} \, dx = 36$

Step-by-step explanation:

The equations are:

$y = x^{2} + 2x + 3$

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The two graphs intersect when:

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To find the area under the curve for the first equation:

$A_{1} = \int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

To find the area under the curve for the second equation:

$A_{2} = \int\limits^3__-3}{2x + 12} \, dx$

To find the total area:

$A = A_{2} -A_{1} = \int\limits^3__-3}{2x + 12} \, dx -\int\limits^3__-3}{x^{2} + 2x + 3} \, dx$

Simplifying the equation:

$A = \int\limits^3__-3}{2x + 12}-({x^{2} + 2x + 3}) \, dx = \int\limits^3__-3}{9}-{x^{2}} \, dx$

Note: The reason the area is equal to the area two minus area one is that the line, area 2, is above the region of interest (see image).

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2 years ago

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3 years ago

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