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2 years ago

The restaurant bill you and your friend received

Mathematics

2 answers:

kramer2 years ago

4 0

Answer:

nice

Step-by-step explanation:

babunello [35]2 years ago

3 0

Answer:

Step-by-step explanation:

tyyyyyyyyyyyyyyyyyyyyyy

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What's 5and24/10 divided by 4?

Alchen [17]

First, convert 5 and 24/10 into a mixed fraction:

5 and 24/10 = 74/10

Now, divide 74/10 by 4:

74/10 ÷ 4 = 74/10 × 1/4
  = 74/40
  = 37/20
  = 1 and 17/20

(Remember that dividing requires you to reciprocate 4)

Hope this helps!

5 0

3 years ago

HELP MEEEEEEEEE PLEASEEEEEEEEEEEEEE!!!!!!!!!!!!!!!!!!

Nadya [2.5K]

Answer: 58, 122

Step-by-step explanation:

$x+2x+6=180\\\\3x+6=180\\\\x+2=60\\\\x=58\\\\\therefore 180-x=122$

4 0

1 year ago

Read 2 more answers

Which equations represent two vertical asymptotes of the function y = 3 cot(4/3^x)?

Maurinko [17]

D ) x = 0 and x = 3 π / 4

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7 0

3 years ago

The expression sin x (cos x cot x - sin x) simplifies to?

nikdorinn [45]

◆ Trigonometric Identities ◆

Hey !!

Check the attachment.
Hope it helps you :)

8 0

3 years ago

Read 2 more answers

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

3 years ago