All categories

2 years ago

Mrs. Rogers sold 49 out of 78 crafts at the fair. What percent of her crafts did she sell? Show your

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

3 0

Answer:

62.8, 63% if you around it off.

Step-by-step explanation: Do 49 Divided by 78, then times it by 100 to get your percentage. I just rounded it off to the nearest tenth and whole number just in case. Hope this helped!

You might be interested in

Write the equation of the line in fully simplified slope-intercept form.

tiny-mole [99]

Step-by-step explanation:

x= -7 y=0

x= -3 y =-6

y =mx+b

b=-7

-6= -3m -7

-3m = 1

m= -1/3

y= -x/3 -7

$y = - \frac{x}{3} - 7$

8 0

2 years ago

Simple equation IS ALSO KNOWN HAS

antoniya [11.8K]

Answer:

linear equation.

Step-by-step explanation:

Please mark as brainliest

6 0

3 years ago

When you draw an image in the right triangle, is it always reflected?

vitfil [10]

No, because the image can be drawn backwards

7 0

2 years ago

Read 2 more answers

mark has a board that is 12 feet long. he cuts the board into 8 pieces that are the same lenghth. how long is each piece?

Anna71 [15]

(12 ft) / 8 = 1.5 ft = 1 foot 6 inches

3 0

3 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$

4 0

1 year ago