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zloy xaker [14]
3 years ago
6

What is the answer to (-15,12),(-9,12)

Mathematics
1 answer:
slava [35]3 years ago
5 0
What do you mean by this? Would you like me to plot it?
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Name the property of multiplication that is shown by the equation 4x· 1 = 4x
Alex_Xolod [135]
Transitive property I believe
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A teacher keeps track of the number of students that participate at least three times in an optional study session each year. He
Anastasy [175]

Answer:

It is not C got it wrong.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the value of this expression if x=5
Nastasia [14]

so, uh, i think the answe is actually 5. Since 5^2 is 25, minus 5 is 20, and 5 minus 1 is 4, 20 divided by 4 is 5. i hope that helps.

4 0
3 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
3 years ago
F(x) = 3x + 6. Find the inverse of f(x).
Levart [38]

Answer:

\tt C) \:\tt f^{-1}(x)=\cfrac{x-6}{3}

Step-by-step explanation:

We're given,

\tt f(x) = 3x + 6

~

\tt y=3x+6

\tt x=3y+6

\tt x-6=3y

\tt y=\cfrac{x-6}{3}

\tt f^{-1}(x)=\cfrac{x-6}{3}

~

3 0
2 years ago
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