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photoshop1234 [79]
3 years ago
6

Fill in the blanks so that the resulting statement is true. If f is a polynomial function and​ f(a) and​ f(b) have opposite​ sig

ns, then there must be at least one value of c between a and b for which ​f(c)equals​_______ This result is called the ​_______ Theorem.
Mathematics
1 answer:
tia_tia [17]3 years ago
6 0

Answer:

f(c) equals zero

Intermediate Value Theorem

Step-by-step explanation:

Intermediate value theorem is one which states that f is a continuous function whose domain contains intervals which are a and b. It takes on any value between f(a) and f(b) at some point within interval. If we know the two values, we can pick any number between those two values and determine its function.

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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
2 years ago
Question 2 Mr. Smith is buying two types of gift cards to give as prizes to employees at a company picnic. He will buy restauran
ella [17]

Answer: He can buy 5 restaurant gift cards and 7 movie gift cards

Step-by-step explanation:

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2 years ago
Find the midpoint of the segment with the given endpoints . G(-3,-5) and H(1,-1)
Serhud [2]
All you have to do for this is follow the midpoint formula.

Let me know if the picture is unclear and I'll type it out for you.

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Find the slope of the line through the<br> given points: (2,6) &amp; (5, 6)
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6-6/5-2 = 0/3

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