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2 years ago

10

I need someone to hire for geometry work i have tons of questions every 10 u complete u get paid 3 dollars dm me if intrested as

ap before post taken down thanks.

1 answer:

IrinaK [193]2 years ago

6 0

Can't you use a calculator?

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A used motorcycle is on sale for $3,600. Erik makes an offer equal to 3/4 of this price. How much does Erik offer for the motorc

lesantik [10]

3/4 of 3600..." of " means multiply
3/4 * 3600 =
10800/4 =
2700 <==

4 0

2 years ago

Read 2 more answers

I need this fast thank you! :)

Alex Ar [27]

Answer:

C

Step-by-step explanation:

Hope this helps! Good Luck!

3 0

2 years ago

Read 2 more answers

Last summers, students who worked earned an average of $80 a week with a standard deviation of $12. What percentage of students

Roman55 [17]

Answer:

2.28%

Step-by-step explanation:

<u>Given:</u>

μ = 80
σ = 12
x = 104

<u>Find z-score:</u>

z = (x - μ)/σ
z = (104 - 80)/12 = 2

Percent value corresponding to 2 as per z-table is 0.9772. This is corresponding to 97.72%.

<u>Percentage of students earned more than $104:</u>

100 - 97.72 = 2.28

5 0

2 years ago

A automated car Wash can watch 30 cars in 2 hours how long would it take it to wash 90 cars

inessss [21]

Answer:

30/5=6 is the answer

Step-by-step explanation:

explanation:

120/6=20 hours... I hope I helped :)

4 0

2 years ago

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

2 years ago