Chapter 6: Problem 76
Simplify. Do not use negative exponents in the answer. $$ \left(-4 x^{3}\right)^{0} $$
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Chapter 6: Problem 76
Simplify. Do not use negative exponents in the answer. $$ \left(-4 x^{3}\right)^{0} $$
These are the key concepts you need to understand to accurately answer the question.
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Hooke's law states that the distance \(d\) that a spring is stretched by a hanging object varies directly as the mass \(m\) of the object. If the distance is \(20 \mathrm{cm}\) when the mass is \(3 \mathrm{kg},\) what is the distance when the mass is \(5 \mathrm{kg} ?\)
Let \(f(x)=\frac{4}{x^{2}-1} \quad\) and \(\quad g(x)=\frac{4 x^{2}+8 x+4}{x^{3}-1}\) Find each of the following. a) \((f \cdot g)(x)\) b) \((f / g)(x)\) c) \((g / f)(x)\)
Evaluate $$ 2 x-y^{2} \div 3 x \text { for } x=3 \text { and } y=-6 $$
The intensity I of light from a bulb varies directly as the wattage of the bulb and inversely as the square of the distance \(d\) from the bulb. If the wattage of a light source and its distance from reading matter are both doubled, how does the intensity change?
The formula $$a=\frac{\frac{d_{4}-d_{3}}{t_{4}-t_{3}}-\frac{d_{2}-d_{1}}{t_{2}-t_{1}}}{t_{4}-t_{2}}$$ can be used to approximate average acceleration, where the \(d^{\prime}\) 's are distances and the \(t^{\prime}\) s are the corresponding times. Solve for \(t_{1}\).
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