Advanced Placement (AP), 17.11.2020 08:50 afitzgerald

1. Describe a time when you changed the quantity demanded for a product or service in response to a change is price.
2. Describe a time when you changed the demand for a product or
service in response to a non-price determinant of demand (TRIBE).

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Answer from: terrysizemore666

See explanation

Explanation:

Demand means the amount of goods and services that a particular individual wants to purchase at a particular price and at a particular place.

Factors which affect demand include price. When price increases, the demand for a good or service reduces and vice versa.

Other price factors that affect demand include taste, fashion, weather, consumer's income etc.

1. There was a particular time that I wanted to buy a pen which costs $2 each. I wanted to buy 5 pens which would be $10. When I got to the stationary store, I was informed that the pen was now $2.50 each. Due to this, I could only buy 4 pen. This shows that due to an increase in price, the quantity demanded of pen that I could buy was reduced.

2. My pay at my workplace got increased from $980 to $1100. Due to these increase in my income, I increased the quantity demand for my food items as I could now buy more food items because my income has risen.

Answer from: Quest

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

Answer from: Quest

This is called visual imagery. it is a memorization technique wherein an idea or a term is associated with an image, this in remembering ideas, especially as humans are very visual learners. this is also why visual aids a lot for memory retention.

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