<span>w(w+5)=20,000 will help you find the answer</span>
Answer:
x=3, solution is not extraneous
Step-by-step explanation:
we have
Solve for x
squared both sides
<u><em>Verify</em></u>
For x=3
-----> is true
therefore
x=3, solution is not extraneous
1/2 divided by 1/6 is 3
use KCF strategy
1/2 x 6/1=3
1x6=6
2x1=2
6/2=3
d1 = 51(t)
d2 = 71(8 - t)
d1 + d2 = 508
and now you just have to solve for t.
Strictly speaking, x^2 + 2x + 4 doesn't have solutions; if you want solutions, you must equate <span>x^2 + 2x + 4 to zero:
</span>x^2 + 2x + 4= 0. "Completing the square" seems to be the easiest way to go here:
rewrite x^2 + 2x + 4 as x^2 + 2x + 1^2 - 1^2 = -4, or
(x+1)^2 = -3
or x+1 =i*(plus or minus sqrt(3))
or x = -1 plus or minus i*sqrt(3)
This problem, like any other quadratic equation, has two roots. Note that the fourth possible answer constitutes one part of the two part solution found above.
