I need to find the area of these shapes

Mathematics

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

1: 374cm

2: 248mm

3: 69m

I believe this is the right answer but I don't think the labels are correct

You might be interested in

What is the shape of the following distribution {3, 4, 4, 5, 5, 5, 8, 8, 8, 7, 7, 7, 2, 6, 6, 6, 15}

bulgar [2K]

Tell me the answer choices.

3 0

4 years ago

What is 2/5 of a book to 1/2 of an hour as a unit rate?

hjlf

Answer: 4/5 book per hour

Step-by-step explanation: in this question, we are to rewrite (2/5)/(1/2) as unit rate.

To write this as a unit rate, what we need to do is to divide 2/5 by 1/2.

That means 2/5 divided by 1/2

Removing the division and replacing with multiplication will give 2/5 * 2/1 = 4/5 books/hour

6 0

3 years ago

Use the Green's Theorem to calculate the work done by the field [ F (x, y) = -3y^5 i + 5y^2x^3 j ] to move a particle along the

lianna [129]

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\iint_R\left(\frac{\partial(5y^2x^3)}{\partial x}-\frac{\partial(-3y^5)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$\dfrac{\partial(5y^2x^3)}{\partial x}=15y^2x^2$
$\dfrac{\partial(-3y^5)}{\partial y}=-15y^2$

$=\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}15y^2(x^2+1)\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates, the integral is equivalent to

$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}15r^3\sin^2\theta(r^2\cos^2\theta+1)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle15\int_0^{2\pi}\int_0^2\left(\frac{r^5}4\sin^22\theta+r^3\sin^2\theta\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac{15}4\left(\int_0^2r^5\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^22\theta\,\mathrm d\theta\right)+15\left(\int_0^2r^3\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)$
$=100\pi$