All categories

4 years ago

A health club charges a sign

Mathematics

1 answer:

Nadusha1986 [10]4 years ago

7 0

30+25m
Total Price: $280

You might be interested in

How do I solve for x? (the picture has the problem)

max2010maxim [7]

Answer:

x = 21.013155617496

Step-by-step explanation:

we have a right triangle then ,

$\cos \left( 36\right) =\frac{17}{x}$

Then

$x=\frac{17}{\cos \left( 36\right) } = 21.013155617496$

Then

x ≈ 21

5 0

2 years ago

Read 2 more answers

Roger is paid $50 to sell videos of a play to an audience after the play is preformed.

konstantin123 [22]

Answer:

Roger can earn $510 at most.

Step-by-step explanation:

We are given the function

Which gives the earnings of Roger when he sells v videos. Since the play’s audience consists of 230 people and each one buys no more than one video, v can take values from 0 to 230, i.e.

That is the practical domain of E(v)

If Roger is in bad luck and nobody is willing to purchase a video, v=0

If Roger is in a perfectly lucky night and every person from the audience wants to purchase a video, then v=230. It's the practical upper limit since each person can only purchase 1 video

In the above-mentioned case, where v=230, then

Roger can earn $510 at most.

3 0

3 years ago

Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2

LUCKY_DIMON [66]

Make a change of coordinates:

$u(x,y)=xy$
$v(x,y)=\dfrac xy$

The Jacobian for this transformation is

$\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}$

and has a determinant of

$\det\mathbf J=-\dfrac{2x}y$

Note that we need to use the Jacobian in the other direction; that is, we've computed

$\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}$

but we need the Jacobian determinant for the reverse transformation (from $(x,y)$ to $(u,v)$ . To do this, notice that

$\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}$

we need to take the reciprocal of the Jacobian above.

The integral then changes to

$\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv$
$=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2$

8 0

3 years ago

Can u pls find the slope intercept form pls?

Irina18 [472]

Answer:

y = - 7/5x + 18/5

Slope: - 7/5

y - intercept: (0, 18/5)

Hope this helps!!

3 0

2 years ago

PLZ ONLY HELP IF UR 100% SURE !!!

Zigmanuir [339]

1. 48,000

2. 50%

3. 1,500, week 5

6 0

3 years ago