Could someone please help me

The linear equation y = 13x + 80 can be used to model the total cost y (in pounds) for x people attending the greyhound racing e

lesantik [10]

Answer:

Justification of 'Cost of Match , based on attendants' linear equation

Step-by-step explanation:

Yes , this linear equation might hold right in denoting total cost 'y' for 'x' people attending racing event

Given equation : y = 13x + 80

Cost 'y' is the variable dependent on explanatory variable ie no. of race attendants. So, <em>y</em> is on LHS, being determined by <em>x</em> on LHS.

80 is the autonomous constant intercept of equation, denoting fixed setup cost irrespective of no. of attendants. +13x denotes every additional attendant increases cost by 13 units (above the level of fixed cost), this is the positive slope of equation (as variables are directly related)

6 0

3 years ago

Jade types 40 words in 7 minutes. Andrew types 35 words in 6 minutes. Who is the faster typist?

GREYUIT [131]

Andrew types faster than Jade

40/7 = 5.7 words per minute (Jade)

35/6 = 5.8 words per minute (Andrew)

May I have brainliest please? :)

6 0

3 years ago

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Et f(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 5)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 11 tha

nignag [31]

Since the surface is closed, and the vector field is rather complicated, you can use the divergence theorem. The flux of $\mathbf f(x,y,z)$ across $S$ is given by a surface integral, which the divergence theorem asserts is equivalent to a volume integral:

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV$

where $R$ denotes the space with boundary $S$ . We have

$\nabla\cdot\mathbf f(x,y,z)=\dfrac{\partial z\tan^{-1}(y^2)}{\partial x}+\dfrac{\partial z^3\ln(x^2+5)}{\partial y}+\dfrac{\partial z}{\partial z}=0+0+1=1$

So in fact the flux across $S$ happens to be equal (in magnitude) to the volume encased by $S$ .

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\int_{x=-3}^{x=3}\int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}\int_{z=2}^{z=11-x^2-y^2}\mathrm dz\,\mathrm dy\,\mathrm dx$

Convert to cylindrical coordinates, setting

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$

$\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{\zeta=2}^{\zeta=11-r^2}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=3}r(11-r^2-2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=3}(9r-r^3)\,\mathrm dr=\dfrac{81\pi}2$

8 0

3 years ago

PLEASE q5 HELPPPP!!!!

ICE Princess25 [194]

Correct answer for Q5:

t=4

3 0

3 years ago

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Round to 1 decimal place 68.87

lyudmila [28]

Answer: 68.9

Explanation:

6 0

3 years ago