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2 years ago

A number greater than -5 written as a problem

Mathematics

1 answer:

bazaltina [42]2 years ago

7 0

The answer is x > -5

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Differentiate y=x⁴(1-2x⁵)⁶(5-8x³)².

Liula [17]

Step-by-step explanation:

Let $y(x)=f(x)g(x)h(x)$ where

$f(x) = x^4$

$g(x)= (1 -2x^5)^6$

$h(x)= (5 - 8x^3)^2$

so that

$y(x) = x^4(1 -2x^5)^6(5 - 8x^3)^2$

Recall that the derivative of the product of functions is

$y'(x)=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)$

so taking the derivatives of the individual functions, we get

$f'(x) = 4x^3$

$g'(x) = 6(1 - 2x^5)^5(-10x^4)$

$h'(x) = 2(5 - 8x^3)(-24x^2)$

So the derivative of y(x) is given by

$y'(x) = 4x^3(1 -2x^5)^6(5 - 8x^3)^2 + x^4 6(1 -2x^5)^5(-10x^4)(5 - 8x^3)^2 + x^4(1 -2x^5)^6 2(5 - 8x^3)(-24x^2)$

$y'(x) = 4x^3(1 -2x^5)^6(5 - 8x^3)^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 60x^8(1 -2x^5)^5(5 - 8x^3)^2$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:- 48x^6(1 -2x^5)^6 2(5 - 8x^3)$

3 0

2 years ago

3x^2 + 4x = -8<br> solve using quadratic formula

Firdavs [7]

Use the quadratic equation to solve this (image of the quadratic equation is below)

Remember that quadratic functions are set up like so:

$ax^{2} +bx +c = 0$

To make the equation 3x² + 4x = -8 into a quadratic function you must bring -8 to the left side of the equation so it equals zero. To do this add 8 to both sides

3x² + 4x + 8= -8 + 8

3x² + 4x + 8 = 0

That means that in this equation...

a = 3

b = 4

c = 8

^^^Plug these numbers into the quadratic equation and solve (Keep in mind that +/- is ± )

$\frac{-4+/-\sqrt{(4)^{2}-4(3)(8)}}{2(3)}$

$\frac{-4+/-\sqrt{16-96}}{6}$

$\frac{-4+/-\sqrt{-80}}{6}$

^^^There is no real solution to this equation. You must have an imaginary solution

$\frac{-4+/-i\sqrt{80}}{6}$

Simplify

$\frac{-2+/-2i \sqrt{5}}{3}$

Hope this helped!

~Just a girl in love with Shawn Mendes

5 0

2 years ago