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BlackZzzverrR [31]
2 years ago
9

Find the divergence of the vector field. v(x, y, z) = 2yzi + 7xzj + 6xyk div(v) = incorrect: your answer is incorrect.

Mathematics
1 answer:
Molodets [167]2 years ago
8 0
\mathrm{div}(\mathbf v)=\dfrac{\partial(2yz)}{\partial x}+\dfrac{\partial(7xz)}{\partial y}+\dfrac{\partial(6xy)}{\partial z}=0
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Find the vertices and foci of the hyperbola. 9x2 − y2 − 36x − 4y + 23 = 0
Xelga [282]
Hey there, hope I can help!

NOTE: Look at the image/images for useful tips
\left(h+c,\:k\right),\:\left(h-c,\:k\right)

\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1\:\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing:H}
with the center of (h, k), semi-axis a and semi-conjugate - axis b.
NOTE: H = hyperbola

9x^2-y^2-36x-4y+23=0 \ \textgreater \  \mathrm{Subtract\:}23\mathrm{\:from\:both\:sides}
9x^2-36x-4y-y^2=-23

\mathrm{Factor\:out\:coefficient\:of\:square\:terms}
9\left(x^2-4x\right)-\left(y^2+4y\right)=-23

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}1
\frac{1}{1}\left(x^2-4x\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x^2-4x+4\right)-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y^2+4y+4\right)=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Convert\:to\:square\:form}
\frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right)

\mathrm{Refine\:}-\frac{23}{9}+\frac{1}{1}\left(4\right)-\frac{1}{9}\left(4\right) \ \textgreater \  \frac{1}{1}\left(x-2\right)^2-\frac{1}{9}\left(y+2\right)^2=1 \ \textgreater \  Refine
\frac{\left(x-2\right)^2}{1}-\frac{\left(y+2\right)^2}{9}=1

Now rewrite in hyperbola standardform
\frac{\left(x-2\right)^2}{1^2}-\frac{\left(y-\left(-2\right)\right)^2}{3^2}=1

\mathrm{Therefore\:Hyperbola\:properties\:are:}\left(h,\:k\right)=\left(2,\:-2\right),\:a=1,\:b=3
\left(2+c,\:-2\right),\:\left(2-c,\:-2\right)

Now we must compute c
\sqrt{1^2+3^2} \ \textgreater \  \mathrm{Apply\:rule}\:1^a=1 \ \textgreater \  1^2 = 1 \ \textgreater \  \sqrt{1+3^2}

3^2 = 9 \ \textgreater \  \sqrt{1+9} \ \textgreater \  \sqrt{10}

Therefore the hyperbola foci is at \left(2+\sqrt{10},\:-2\right),\:\left(2-\sqrt{10},\:-2\right)

For the vertices we have \left(2+1,\:-2\right),\:\left(2-1,\:-2\right)

Simply refine it
\left(3,\:-2\right),\:\left(1,\:-2\right)
Therefore the listed coordinates above are our vertices

Hope this helps!

8 0
3 years ago
Use a table to find a z-score that fits the given conditions. Interpolate if necessary.
kotegsom [21]
<span>After deeply looking at the table it seems 
z= .05
which means only 52% is less than Z

so, that is greater than z by 48%</span>
7 0
3 years ago
What number is nine<br> tenths more than<br> 0.852
kirill115 [55]

Answer:

1.752

Step-by-step explanation:

Nine tenths = 0.9

0.852 + 0.9 = 1.752

4 0
3 years ago
Select two choices that are true about the function f(x)=23x+14/x
WINSTONCH [101]

Answer:

There is an asymptote at x = 0

There is an asymptote at y = 23

Step-by-step explanation:

Given the function:

(23x+14)/x

Vertical asymptote is gotten by equating the denominator to zero

Since the denominator is x, hence the vertical asymptote is at x = 0. This shows that there is an asymptote at x = 0

Also for the horizontal asymptote, we will take the ratio of the coefficient of the variables in the numerator and denominator

Coefficient of  x at the numerator = 23

Coefficient of x at the denominator = 1

Ratio = 23/1 = 23

This means that there is an asymptote at y = 23

8 0
2 years ago
The height of a triangle is two more than three times the base. Determine the dimensions that will give a total area of 28 yards
jonny [76]
H = 3b+2
A = (h*b)/2     28 = (3b+2)b/2     56 = 3b²+2b    0 = 3b² + 2b - 56

⊕\left \{ {{y=2} \atop {x=2}} \right.  \int\limits^a_b {x} \, dx  \lim_{n \to \infty} a_n   \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]  \beta  \\  \\  \\  x^{2}  \sqrt{x}  \sqrt[n]{x}  \frac{x}{y}  x_{123}  x^{123}  \leq  \geq  \pi  \alpha   \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]   \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]  x_{123}  \int\limits^a_b {x} \, dx  \left \{ {{y=2} \atop {x=2}}
ω
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∩
8 0
3 years ago
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