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abruzzese [7]
2 years ago
12

If y varies inversely as x and y=23 when x=8 find y when x=4

Mathematics
1 answer:
valina [46]2 years ago
5 0
<h2>INVERSE VARIATION</h2>

<em>E</em><em>Q</em><em>U</em><em>A</em><em>T</em><em>I</em><em>O</em><em>N</em><em>:</em><em> </em><em> </em><em> </em><em> </em>

y =  \frac{k}{x}

<em>–</em><em>–</em><em>–</em><em>–</em><em>–</em><em>–</em><em>–</em><em>–</em>

First, find the constant of variation <em>k</em> using the formula k=xy.

23 =  \frac{k}{8}

k = (8)(23)

k = 184

<em>–</em><em>–––––––</em>

Then, find <em>y</em> if <em>x</em> is equal to four using the formula y=184/x.

y =  \frac{184}{x}

y =  \frac{184}{4}

y = 46

<em>F</em><em>i</em><em>n</em><em>a</em><em>l</em><em> </em><em>A</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em>:</em>

<h2>y = 46</h2>
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Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

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You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

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Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

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Answer:

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Zielflug [23.3K]

Answer:

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Answer:

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