√96 how do i solve this and it can't be a decimal

1 answer:

8 0

4 square root 6 :)

It would be the square root of 16*6 and the square root of 16 is 4

So in conclusion 4 square root sign 6

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(07.08)

Black_prince [1.1K]

If the given expression is simplified we get a. -4x² - 3x + 2.

Explanation:

The numerator has three terms while the denominator only has one term. We divide the numerator terms each separately with the denominator term.

So 8x³ + 6x² - 4x / -2x becomes (8x³ / -2x) + (6x² / -2x) + (- 4x/ -2x).

The simplification of the first term; 8 / -2 = -4, x³ / x = x². So the first term is -4x².

The simplification of the second term; 6 / -2 = -3, x² / x = x. So the second term is -3x.

The simplification of the third term; -4 / -2 = 2, x / x = 1. So the third term is 2.

Adding all the terms we get -4x² - 3x + 2. This is the option a.

6 0

4 years ago

Find dy/dx for 4 - xy = y^3

storchak [24]

Answer:

$\frac{dy}{dx}=-\frac{y}{3y^2+x}$

Step-by-step explanation:

4-xy=y^3

dy/dx=?

$\frac{d(4-xy)}{dx}=\frac{d(y^3)}{dx}\\ \frac{d(4)}{dx}-\frac{d(xy)}{dx}=3y^{3-1}\frac{dy}{dx}\\ 0-(\frac{dx}{dx}y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(1y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -(y+x\frac{dy}{dx})=3y^2\frac{dy}{dx}\\ -y-x\frac{dy}{dx}=3y^2\frac{dy}{dx}$