Answer:
x = 1
Step-by-step explanation:
For a quadratic function 
, the axis of symmetry is
.
This comes from using part of the Quadratic Formula for finding zeros of the function:
See the -b/(2a) in the first term? That's the x-coordinate of the vertex (turning point) of the parabolic graph of the function. The axis of symmetry goes through the vertex.
Answer:
$3.50
Step-by-step explanation:
2.5% x 25 = 62.5%
2.50 divide 62.5% = 4
4x25=1.00
2.50+1.00=3.50
I think it's D. The area of A is k2 times the area of rectangle B
The first time I did it, I got an answer that's not one of the choices. The second time
I did it, I got an answer that IS . Here are both of my procedures. If all you want is
the answer, look down below at the second one. But if you could help me out, now
that you know how to do this stuff, please look at my first solution and tell me where
I messed up. I can't find it.
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Here's what the problem tells you:
D = 10 log ( ' I ' / 10⁻¹² )
D = 60 . . . . . find ' I ' .
Here we go:
60 = 10 log ( ' I ' / 10⁻¹² )
Divide each side by 10 :
6 = log ( ' I ' / 10⁻¹² )
Raise 10 to the power of each side of the equation:
10⁶ = ' I ' / 10⁻¹²
Multiply each side by 10¹² :
10¹⁸ = ' I ' That's 10^18. It looks bad, because that isn't one of the choices.
Let's try a slightly different procedure:
============================================
After substituting 10⁺¹² for I₀ , we're working with this formula:
D = 10 log ( 'I' / 10⁺¹² )
Let's just look at the log part of that.
The log of a fraction is [ log(numerator) - log(denominator) ]
log of this fraction is [ log( 'I' ) - log(10⁻¹²) ]
But log(10⁻¹²) is just (-12) .
So the log of the fraction is [ log( 'I' ) + 12 ]
And the whole formula is now:
D = 10 [ log( 'I' ) + 12 ]
60 = 10 [ log( 'I' ) + 12 ]
Divide each side by 10 :
6 = log( 'I' ) + 12
Subtract 12 from each side :
-6 = log ( ' I ' )
' I ' = 10⁻⁶
That's choice-'B' .
==================================================
I'm going to leave the first solution up there, in hopes that you, or one
of the many aces, experts, and geniuses that prowl this site constantly,
can weigh in and show me my blunder on the first attempt.
