Ask question

Ask question

All categories

3 years ago

14

Megan and her little sister, Nora, counted their Halloween candy after trick-or-treating. They love taffy and wanted to know who

collected more taffy pieces compared to the rest of the candy. In Megan's bag, 20 out of the 50 candy pieces were taffy. In Nora's bag, 11 out of the 25 pieces were taffy. Who got the greater ratio of taffy to total candy pieces?
A.
Megan got the greater ratio
B.
Nora got the greater ration.
C.
neither. the girls got the same ration

1 answer:

ludmilkaskok [199]3 years ago

6 0

Answer:

A. Megan got the greater ratio

Step-by-step explanation:

the numbers are bigger than Nora's numbers

You might be interested in

Which sentence is correctly punctuated?

Mnenie [13.5K]

Answer:

C

Step-by-step explanation:

Because it has a coma after example so that means a pause hope this helps.

5 0

3 years ago

Read 2 more answers

Please hellppp please hellpp

Solnce55 [7]

Answer:

- 3 and 5

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 3x + 5 ← is in slope- intercept form

with slope m = - 3 and y- intercept c = 5

4 0

3 years ago

Read 2 more answers

Suppose a > 0 is constant and consider the parameteric surface sigma given by r(phi, theta) = a sin(phi) cos(theta)i + a sin(

Gnom [1K]

$\Sigma$ should have parameterization

$\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k$

if it's supposed to capture the sphere of radius $a$ centered at the origin. ( $\sin\theta$ is missing from the second component)

a. You should substitute $x=a\sin\varphi\cos\theta$ (missing $\cos\theta$ this time...). Then

$x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2$

as required.

b. We have

$\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k$

$\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath$

$\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k$

$\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi$

c. The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi$

You don't need a substitution to compute this. The integration limits are constant, so you can separate the variables to get two integrals. You'd end up with

$\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2$

# # #

Looks like there's an altogether different question being asked now. Parameterize $\Sigma$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $\sqrt2\le u\le\sqrt6$ and $0\le v\le2\pi$ . Then

$\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}$

The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The integrand doesn't depend on $v$ , so integration with respect to $v$ contributes a factor of $2\pi$ . Substitute $w=1+4u^2$ to get $\mathrm dw=8u\,\mathrm du$ . Then

$\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3$

# # #

Looks like yet another different question. No figure was included in your post, so I'll assume the normal vector points outward from the surface, away from the origin.

Parameterize $\Sigma$ by

$\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k$

with $-1\le u\le1$ and $0\le v\le 2$ . Take the normal vector to $\Sigma$ to be

$\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath$

Then the flux of $\vec F$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43$

If instead the direction is toward the origin, the flux would be positive.

8 0

4 years ago

Using the number line above, write the integer that each point represents

zloy xaker [14]

Answer:

E= -6

D= 4

B =- 4

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The number multiplied by the variable is called the ______________ of the variable.

nirvana33 [79]

Answer:

coefficient

explanation:

i just know the answer

6 0

3 years ago