Let the second score = x
⇒ second score = x
<span>the first is 14 points more than the second
</span>⇒ first score = x + 14
<span>the sum of the first two is 6 more than twice the third
</span>⇒ third score = 1/2 (x + x + 14 - 6) = x + 4
<span>The sum of a student's three score is 246
</span>⇒ x + (x + 14) + (x + 4) = 246
<span>
Solve x:
</span>x + x + 14 + x + 4 = 246
3x + 18 = 246
3x = 228
x = 76
second score = x = 76
first score = x + 14 = 76 + 14 = 90
Answer: 90
Uh... 3 1/2 jail? I don't know.
<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>
