answer:The derivative of the function ( f(x) ) is given by: [ f'(x) = frac{d}{dx}[log (-4x - 5)] + frac{d}{dx}[tanh (9 - 4x)] ] Using the derivative rules for logarithms and the hyperbolic tangent function, we get: [ f'(x) = frac{-4}{-4x - 5} - 4 text{sech}^2(9 - 4x) ] Simplifying the first term: [ f'(x) = frac{4}{4x + 5} - 4 text{sech}^2(9 - 4x) ]

answer:180,000

answer:To multiply these polynomials, we use the distributive property (also known as FOIL for binomials, which stands for First, Outer, Inner, Last). Here's the expansion step by step: [ p(x) cdot q(x) = (4x^2)(7x^2) + (4x^2)(x) + (4x^2)(-2) + (9x)(7x^2) + (9x)(x) + (9x)(-2) + (3)(7x^2) + (3)(x) + (3)(-2) ] Now, we simplify each term: [ = 28x^4 + 4x^3 - 8x^2 + 63x^3 + 9x^2 - 18x + 21x^2 + 3x - 6 ] Combine like terms: [ = 28x^4 + (4x^3 + 63x^3) + (-8x^2 + 9x^2 + 21x^2) + (-18x + 3x) - 6 ] [ = 28x^4 + 67x^3 + 22x^2 - 15x - 6 ] So, the expanded product is 28x^4 + 67x^3 + 22x^2 - 15x - 6.

answer:Resistors R1 and R2 serve as pull-up and pull-down resistors in the circuit. Their primary role is to ensure that the MOSFET gates assume the appropriate voltage state when there is no input signal. The gate of a MOSFET behaves akin to a capacitor. In this configuration, pressing the ON button connects the gate of Q3 to a higher voltage, charging it and turning the MOSFET on. Upon releasing the button, the gate discharges through R3, which then turns Q3 off. If R3 were absent, the gate would remain charged, keeping Q3 in the on state. A similar mechanism applies to Q1 and R1. Activating either Q2 or Q3 turns on Q1, and R1 guarantees that when both Q2 and Q3 are off, Q1 turns off as well. It's important to note that when the gates are driven by a low-impedance source, the resistors do not significantly affect the Vgs (gate-source voltage) value of the MOSFETs.