Find the volume of this prism

The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.

saw5 [17]

Answer:

$x^2 + x -12$

Step-by-step explanation:

If the solutions are x=-4 and x=3, then it must have factors (x+4)(x-3). Assuming there is no GCF or leading coefficient other than 1, multiply with FOIL to find the standard form.

$(x+4)(x-3)\\x^2 + 4x -3x -12\\x^2 + x -12$

7 0

2 years ago

Read 2 more answers

37. Verify Green's theorem in the plane for f (3x2- 8y2) dx + (4y - 6xy) dy, where C is the boundary of the

Nastasia [14]

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

• (38) was addressed in 24438105

• (39) was addressed in 24434477

• (40) and (41) were both addressed in 24434541

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

$D = \left\{(x,y) \mid 0\le x\le1 \text{ and }x^2\le y\le x\right\}$

• Compute the line integral directly by splitting up C into two component curves,

C₁ : x = t and y = t ² with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = 1 - t with 0 ≤ t ≤ 1

Then

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

*** 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:

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

(b) C is the boundary of the region

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

• Compute the line integral directly, splitting up C into 3 components,

C₁ : x = t and y = 0 with 0 ≤ t ≤ 1

C₂ : x = 1 - t and y = t with 0 ≤ t ≤ 1

C₃ : x = 0 and y = 1 - t with 0 ≤ t ≤ 1

Then

$\displaystyle \int_C = \int_{C_1} + \int_{C_2} + \int_{C_3} \\\\ = \int_0^1 3t^2\,\mathrm dt + \int_0^1 (11t^2+4t-3)\,\mathrm dt + \int_0^1(4t-4)\,\mathrm dt \\\\ = \int_0^1 (14t^2+8t-7)\,\mathrm dt = \boxed{\frac53}$

• Using Green's theorem:

$\displaystyle \int_C (3x^2-8y^2)\,\mathrm dx + (4y-6xy)\,\mathrm dx = \int_0^1\int_0^{1-x}10y\,\mathrm dy\,\mathrm dx = \boxed{\frac53}$

4 0

2 years ago

Clevon forms a triangle with three straws measuring 20, 25, and 18 inches. To the nearest whole

Ne4ueva [31]

Answer:

Step-by-step explanation:

Let's label this triangle as triangle ABC. Side AB is 18, side BC is 20 and side CA is 25 and the angle we are looking for is angle C. Use the Law of Cosines to find the missing angle. You have to use the Law of Cosines because in order to use the Law of Sines you have to have an angle given and we don't so we have no other options. In our case,

$c^2=a^2+b^2-2abcosC$ which for us looks like this: