How to solve (4b)/8 + (2b)/8 =

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 (4b)/8 + (2b)/8 = | Equatix

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Question

$$\frac{4b}{8}+\frac{2b}{8}=$$

Evaluate

$\frac{3b}{4}$

Solution Steps

Divide $4b$ by $8$ to get $\frac{1}{2}b$.

$$\frac{1}{2}b+\frac{2b}{8}$$

Divide $2b$ by $8$ to get $\frac{1}{4}b$.

$$\frac{1}{2}b+\frac{1}{4}b$$

Combine $\frac{1}{2}b$ and $\frac{1}{4}b$ to get $\frac{3}{4}b$.

$$\frac{3}{4}b$$

Show Solution

Differentiate w.r.t. b

$\frac{3}{4} = 0.75$

Steps Using Definition of a Derivative

Divide $4b$ by $8$ to get $\frac{1}{2}b$.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{1}{2}b+\frac{2b}{8})$$

Divide $2b$ by $8$ to get $\frac{1}{4}b$.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{1}{2}b+\frac{1}{4}b)$$

Combine $\frac{1}{2}b$ and $\frac{1}{4}b$ to get $\frac{3}{4}b$.

$$\frac{\mathrm{d}}{\mathrm{d}b}(\frac{3}{4}b)$$

The derivative of $ax^{n}$ is $nax^{n-1}$.

$$\frac{3}{4}b^{1-1}$$

Subtract $1$ from $1$.

$$\frac{3}{4}b^{0}$$

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

$$\frac{3}{4}\times 1$$

For any term $t$, $t\times 1=t$ and $1t=t$.

$$\frac{3}{4}$$