All categories

3 years ago

Please! I Need Help!

1 answer:

6 0

The sales tax is 6.7% from $498.00 -> The sales tax = 498 x 6.7% = $33.366.

The total cost of the trampoline = 498 + 33.366 = $531.366

You might be interested in

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

FIVE STARS AND BRAINLIEST TO CORRECT ANSWER

spin [16.1K]

The series is -10-2+6+14+...+110
We see that the common ratio is 8, so it is of the form T0+8n
We also see that T0=-10 (i.e. when the pattern starts.
So we have
Tn=-10+8n
and the summation would be
∑ (-10+8n)

Next to determine the limits.
We know that the first term is -10, which fits T0=-10+8(0).
The last term is 110, which gives the equation Tn=110=-10+8(n)
Solving for n gives
n=(110+10)/8=15

Therefore the limits of summation are 0, 15
hence the summation formula is
$\sum_0^{15}(-10+8n)$

I will leave the other problem you posted for your own exercise.

3 0

3 years ago

Which ordered pair is a solution to the system of inequalities?

barxatty [35]

Answer:

Step-by-step explanation:

To determine which ordered pair is a solution.

Substitute the x and y values into the inequalities.

Note that both must be true for the pair to be a solution of the system.

(4, 8)

8 > 2(4) → 8 > 8 ← False

8 > 7 ← True

(0, 0)

0 > 2(0) → 0 > 0 ← False

0 > 7 ← False

(3, 7)

7 > 2(3) → 7 > 6 ← True

7 > 7 ← False

(1, 9)

9 > 2(1) → 9 > 2 ← True

9 > 7 ← True

Thus (1, 9) is a solution to the system of equations. → D

6 0

3 years ago

HELP ME PLS I DONT HAVE TIME

chubhunter [2.5K]

Answer:

A is true.......................b is false....................and c is false

Step-by-step explanation:

6 0

3 years ago

Write an equation for the given sentence. Three more than five times x is 38

Margaret [11]

Answer:

3 + 5x = 38

Three more than five times equals 38

6 0

4 years ago