How to solve 5+3+4+3+5-x630

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 5+3+4+3+5-x630 | Equatix

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Question

$$5+3+4+3+5-x630$$

Evaluate

$20-x_{630}$

Solution Steps

Add $5$ and $3$ to get $8$.

$$8+4+3+5-x_{630}$$

Add $8$ and $4$ to get $12$.

$$12+3+5-x_{630}$$

Add $12$ and $3$ to get $15$.

$$15+5-x_{630}$$

Add $15$ and $5$ to get $20$.

$$20-x_{630}$$

Show Solution

Differentiate w.r.t. x_630

$-1$

Steps Using Derivative Rule for Sum

Add $5$ and $3$ to get $8$.

$$\frac{\mathrm{d}}{\mathrm{d}x_{630}}(8+4+3+5-x_{630})$$

Add $8$ and $4$ to get $12$.

$$\frac{\mathrm{d}}{\mathrm{d}x_{630}}(12+3+5-x_{630})$$

Add $12$ and $3$ to get $15$.

$$\frac{\mathrm{d}}{\mathrm{d}x_{630}}(15+5-x_{630})$$

Add $15$ and $5$ to get $20$.

$$\frac{\mathrm{d}}{\mathrm{d}x_{630}}(20-x_{630})$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is $0$. The derivative of $ax^{n}$ is $nax^{n-1}$.

$$-x_{630}^{1-1}$$

Subtract $1$ from $1$.

$$-x_{630}^{0}$$

For any term $t$ except $0$, $t^{0}=1$.

$$-1$$