Solve -7.32*40.16+(-0.16)*(-7.32)+7.32x-160

$$- 7.32 \times 40.16 + ( - 0.16 ) \times ( - 7.32 ) + 7.32 x - 160$$

Evaluate

$$\frac{183x}{25}-452.8$$

Solution Steps

Multiply -7.32 and 40.16 to get -293.9712.

$$-293.9712-0.16\left(-7.32\right)+7.32x-160$$

Multiply -0.16 and -7.32 to get 1.1712.

$$-293.9712+1.1712+7.32x-160$$

Add -293.9712 and 1.1712 to get -292.8.

$$-292.8+7.32x-160$$

Subtract 160 from -292.8 to get -452.8.

$$-452.8+7.32x$$

Differentiate w.r.t. x

$$7.32$$

Steps Using Derivative Rule for Sum

Multiply -7.32 and 40.16 to get -293.9712.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-293.9712-0.16\left(-7.32\right)+7.32x-160)$$

Multiply -0.16 and -7.32 to get 1.1712.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-293.9712+1.1712+7.32x-160)$$

Add -293.9712 and 1.1712 to get -292.8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-292.8+7.32x-160)$$

Subtract 160 from -292.8 to get -452.8.

$$\frac{\mathrm{d}}{\mathrm{d}x}(-452.8+7.32x)$$

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}.

$$7.32x^{1-1}$$

Subtract 1 from 1.

$$7.32x^{0}$$

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

$$7.32\times 1$$

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

$$7.32$$