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3 years ago

A plumber charges $17 to visit a client's home, plus an additional $36 per hour. What is the

Mathematics

2 answers:

jarptica [38.1K]3 years ago

4 0

Because this is a linear problem, I'd use the slope-intercept formula:

$y = mx + b$

where

m = slope (rise/run)

y = dependent value

x = independent value

b = constant value

Because the plumber charges $17 upon arrival, it is our constant. The $36 is our slope (per hour, per minute, per etc. is generally a give away). Additionally because the total cost of the plumber's visit depends on how long s/he stays at a client's home, the $89 is our y value. Therefore the equation should look like this:

$89 = 36x + 17$

Solve for x by subtracting 17 from both sides of the equation then dividing by 36.

$72 = 36x$

$2 = x$

The plumber spent two hours at his/her client's home.

larisa86 [58]3 years ago

3 0

Answer:

I believe the duration for $89.00 is 2 hours

Step-by-step explanation:

We start off with the $17 dollars and calculate 36.00+36.00 for a total of $89. Therefore, (120 mins)/60=2 hours.

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ipn [44]

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CaHeK987 [17]

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2 years ago

Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

sashaice [31]

Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$