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nikitadnepr [17]
3 years ago
6

Find an equation of variation in which y varies jointly as x and z and inversely as the product of w and​ p, where yequalsstartf

raction 7 over 28 endfraction when xequals7​, zequals4​, wequals7​, and pequals8.
Mathematics
1 answer:
leva [86]3 years ago
7 0

To solve this problem you must apply the proccedure shown below:

1. You have that y varies jointly as x and z and inversely as the product of w and​ p. Therefore, you can write the following equation, where k is the constant of proportionality:

y=k(\frac{xz}{wp} )

2. Now, you must solve for the constant of proportionality, as following:

k=\frac{ywp}{xz}

3. Susbtiute values:

y=\frac{7}{28} \\ x=7\\ z=4\\ w=7\\ p=8

k=\frac{(\frac{7}{28})(7)(8))}{(7)(4)}  =0.5

4. Substitute the value of the constant of proportionality into the equation:

y=0.5(\frac{xz}{wp})

The answer is: y=0.5(\frac{xz}{wp})

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Answer:

\displaystyle 36\:units

Step-by-step explanation:

Split up this Isosceles, Right Triangle into two congruent smaller right triangles. The reflexive side [the line that splits them apart] is 6 units, and both legs are 8 units, leaving the hypotenuses to AUTOMATICALLY be 10 units, according to the Pythagorean Theorem:

\displaystyle a^2 + b^2 = c^2 \\ \\ 6^2 + 8^2 = 10^2 \\ \\ 36 + 64 = 100 \\ \\ 100 = 100

With this Pythagorean Triple, we know that our dimensions are correct. Now, to find the perimeter, just add up all the sides EXCEPT for the divider:

\displaystyle 2[10] + 2[8] = 20 + 16 = 36

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Use the divergence theorem to find the outward flux of the vector field F(x,y,z)=2x2i+5y2j+3z2k across the boundary of the recta
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Answer:

The answer is "120".

Step-by-step explanation:

Given values:

F(x,y,z)=2x^2i+5y^2j+3z^2k \\

differentiate the above value:

div F =2 \frac{x^2i}{\partial x}+5 \frac{y^2j}{\partial y}+3 \frac{z^2k }{\partial z}  \\

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\ flu \ of \ x = \int  \int div F dx

              = \int\limits^1_0 \int\limits^3_0 \int\limits^1_0 {(4x+10y+6z)} \, dx \, dy \, dz \\\\ = \int\limits^1_0 \int\limits^3_0 {(4xz+10yz+6z^2)}^{1}_{0} \, dx \, dy  \\\\ = \int\limits^1_0 \int\limits^3_0 {(4x+10y+6)} \, dx \, dy  \\\\ = \int\limits^1_0  {(4xy+10y^2+6y)}^3_{0} \, dx   \\\\ = \int\limits^1_0  {(12x+90+18)}\, dx   \\\\= {(12x^2+90x+18x)}^{1}_{0}   \\\\= {(12+90+18)}   \\\\=30+90\\\\= 120

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