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2 years ago

My geometry teacher doesn't really teach so I need help!

Mathematics

1 answer:

stepladder [879]2 years ago

3 0

Answer:

g(3) = 6

Step-by-step explanation:

The right side of the equation is the part with all the information for the doing of the math.

x^2 - 3 means square the number and then subtract 3.

On the left side of the equation is the g(x). g(x) or f(x) or k(x) is literally just y. Someone could ask what is f(x)? It is y.

What is g(x)? It is y.

What is g? It is the name of the function. So instead of saying ..."the thing where you square a number and then subtract 3..." you can just say "g". Now the (x) beside g DOESNOT mean times, bc sometimes parenthesis means times. THIS IS NOT TIMES!

The (x) is like a billboard. Tells you who will be starring as x in the work on the right side of the equation.

If you see g(x)=x^2-3

then whatever is in the ( ) on the left, put that in the place of x on the right.

g() = ^2 - 3

g(12) = 12^2 - 3

g(-7) = (-7)^2 - 3

or like your question

g(3) = 3^2 - 3

do the calculation

g(3) = 9 - 3

g(3) = 6

done! Hope this helps!

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I'll only look at (37) here, since

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• (39) was addressed in 24434477

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In both parts, we're considering the line integral

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy$

and I assume C has a positive orientation in both cases

(a) It looks like the region has the curves y = x and y = x ² as its boundary***, so that the interior of C is the set D given by

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$\displaystyle \int_C = \int_{C_1} + \int_{C_2} \\\\ = \int_0^1 \left((3t^2-8t^4)+(4t^2-6t^3)(2t))\right)\,\mathrm dt \\+ \int_0^1 \left((-5(1-t)^2)(-1)+(4(1-t)-6(1-t)^2)(-1)\right)\,\mathrm dt \\\\ = \int_0^1 (7-18t+14t^2+8t^3-20t^4)\,\mathrm dt = \boxed{\frac23}$

*** Obviously this interpretation is incorrect if the solution is supposed to be 3/2, so make the appropriate adjustment when you work this out for yourself.

• Compute the same integral using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dy = \iint_D \frac{\partial(4y-6xy)}{\partial x} - \frac{\partial(3x^2-8y^2)}{\partial y}\,\mathrm dx\,\mathrm dy \\\\ = \int_0^1\int_{x^2}^x 10y\,\mathrm dy\,\mathrm dx = \boxed{\frac23}$

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$D = \left\{(x,y) \mid 0\le x\le 1\text{ and }0\le y\le1-x\right\}$

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