F'(x)=-2x{e^(-x²)}
Equating f'(x)=0,
i.e,
-2x{e^(-x²)}=0....eqn(i)
For maximum slope:
Differentiating (i) wrt x,we get,
4x² {e^(-x²)}-2 {e^(-x²)}=0
{e^(-x²)} [4x²-2]=0
4x²-2=0
x=1/√2
If f(x)=y then f'(x)=dy/dx=slope
So,
Maximum slope
= f'(x)
= -2x{e^(-x²)}
=-2×(1/✓2){e^(-1/✓2)²}
= -✓2{e^(-1/2)}
=(-sqrt 2/sqrt e)
=-sqrt(2/e)➡c is correct
Answer:
Because for the first problem you would have to simplify 8 divided by 2 into 4 in which 12 - 4 = 8 then you would have to simplify 12 - 4 to 8 which 8 = 8
Because for the second problem you would have to simplify 12 - 8 to 4 which 4 divided by 2 = 2 then you would have to simplify 4 divided by 2 to 2 which 2 = 2
There are parentheses which if there weren't you would get a different answer
Hope I helped
Sorry if I didnt
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A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
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prev Statement of a problem № 2618 next
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
main
prev Statement of a problem № 2618 next
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a load of 8.00 kg is hung on its end. Find the stress, the strain, and the Young’s modulus for the material of the wire.
Solution:
A metal wire 75.0 cm long and 0.130 cm in diameter stretches 0.0350 cm when a lo
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Answer:
832 beats
Step-by-step explanation:
56/7 = 8 beats per measure
104 x 8
832
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