2 years ago

Let y=4x^2 find the differential dy when x=5 and dx=0.4

Mathematics

1 answer:

anzhelika [568]2 years ago

6 0

Dy=y'*dx
=8x*dx
=8(5)*0.4
=40*0.4
=16

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Please help, this will just take a minute of your time ❤︎ (literally)<br>(Discriminant Worksheet)

igomit [66]

Answer:

$\large\boxed{x=-\dfrac{7-\sqrt{97}}{4}\ \vee\ x=-\dfrac{7+\sqrt{97}}{4}}$

Step-by-step explanation:

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta0,\ \text{then an equation has two solutions:}\ x=\dfrac{-b\pm\sqrt\Delta}{2a}\\\\\text{We have:}\\\\-2x^2-7x+10=4\qquad\text{subtract 4 from both sides}\\\\-2x^2-7x+6=0$

$a=-2,\ b=-7,\ c=6\\\\\Delta=(-7)^2-4(-2)(6)=49+48=97>0\\\\\sqrt\Delta=\sqrt{97}\\\\x=\dfrac{-(-7)\pm\sqrt{97}}{2(-2)}=\dfrac{7\pm\sqrt{97}}{-4}=-\dfrac{7\pm\sqrt{97}}{4}$