<span><span> y2(q-4)-c(q-4)</span> </span>Final result :<span> (q - 4) • (y2 - c)
</span>
Step by step solution :<span>Step 1 :</span><span>Equation at the end of step 1 :</span><span><span> ((y2) • (q - 4)) - c • (q - 4)
</span><span> Step 2 :</span></span><span>Equation at the end of step 2 :</span><span> y2 • (q - 4) - c • (q - 4)
</span><span>Step 3 :</span>Pulling out like terms :
<span> 3.1 </span> Pull out q-4
After pulling out, we are left with :
(q-4) • (<span> y2</span> * 1 +( c * (-1) ))
Trying to factor as a Difference of Squares :
<span> 3.2 </span> Factoring: <span> y2-c</span>
Theory : A difference of two perfect squares, <span> A2 - B2 </span>can be factored into <span> (A+B) • (A-B)
</span>Proof :<span> (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 <span>- AB + AB </span>- B2 =
<span> A2 - B2</span>
</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication.
Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.
Check : <span> y2 </span>is the square of <span> y1 </span>
Check :<span> <span> c1 </span> is not a square !!
</span>Ruling : Binomial can not be factored as the difference of two perfect squares
Final result :<span> (q - 4) • (y2 - c)
</span><span>
</span>
Hello,
Vertices are on a line parallele at ox (y=-3)
The hyperbola is horizontal.
Equation is (x-h)²/a²- (y-k)²/b²=1
Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)
Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6
c²=a²+b²==>b²=36-4²=20
Equation is:
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As
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Miss
Ss
Smsms
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SMS
Answer: no it is not
Step-by-step explanation: any number that repeats forever cannot be written as a rational way or a fraction. So it is not rational.
The seats must be arranged in a 38 by 39 arrangement to have the minimum amount of seats used.
You can find this by finding the square root of 1450, which is 38.0788655293.
