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julsineya [31]
4 years ago
9

Solve with solution...no spam answers pls

Mathematics
2 answers:
Agata [3.3K]4 years ago
6 0
<span>Let's analyze Hannah's work, step-by-step, to see if she made any mistakes. 

</span>In Step 1, Hannah wrote \dfrac{d}{dx} (-3+8x) <span> as the sum of two separate derivatives </span>\dfrac{d}{dx}(-3)+ \dfrac{d}{dx} (8x) <span>using the </span><span>sum rule.
</span>
This step is perfectly fine. 

In Step 2, \dfrac{d}{dx}(8x) was kept as it is, and \dfrac{d}{dx}(-3) was rewritten as 0 using the constant rule.Indeed, according to the constant rule, the derivative of a constant number is equal to zero.

This step is perfectly fine. 

In Step 3, \dfrac{d}{dx} (8x)  was rewritten as \dfrac{d}{dx}(8) \dfrac{d}{dx}(x) supposedly using the constant multiple rule.

The problem is that according to the constant multiple rule, \dfrac{d}{dx}(8x)&#10; should be rewritten as 8 \dfrac{d}{dx}(x) and not as \dfrac{d}{dx}(8)\dfrac{d}{dx}(x).  

<span>Therefore, Hannah made a mistake in this step.</span>
djverab [1.8K]4 years ago
6 0
Step 3, it should have been 8*d/dx(x) which leads to

Step 4 = 8*1  which leads to

Step 5 = 8
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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)}

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3 years ago
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katen-ka-za [31]

Step-by-step explanation:

Identify the method that will be used to solve for x for each equation.

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3 years ago
Add 5/8 inch, 11/16 inch, and 27/32 inch.
noname [10]
Hi,

Work:

Equation;

\frac{5}{8} + \frac{11}{16} + \frac{27}{32}

Write all numerators above the least common denominator 32.

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Or (DECIMAL RESULT)

2.15625

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r3t40
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