Answer:
g(0.9) ≈ -2.6
g(1.1) ≈ 0.6
For 1.1 the estimation is a bit too high and for 0.9 it is too low.
Step-by-step explanation:
For values of x near 1 we can estimate g(x) with t(x) = g'(1) (x-1) + g(1). Note that g'(1) = 1²+15 = 16, and for values near one g'(x) is increasing because x² is increasing for positive values. This means that the tangent line t(x) will be above the graph of g, and the estimates we will make are a bit too big for values at the right of 1, like 1.1, and they will be too low for values at the left like 0.9.
For 0.9, we estimate
g(0.9) ≈ 16* (-0.1) -1 = -2.6
g(1.1) ≈ 16* 0.1 -1 = 0.6
Fairly certain the answer is C
Answer:
Step-by-step explanation:
This is not a linear relationship, it doesn't have a set increase or uniform slope and it seems more like an exponential or quadratic curve to me when I graph it. As you can see you cant draw a straight line through the data points so it cannot be a linear relationship.
72 × 1.4666666666667 = 105.6 feet per second
Strength [is proportional to] d^2
strength1/(d1)^2 = strength2/(d2)^2
800kg/(2cm)^2 = strength2/(3cm)^2
strength2 = 800 kg * (3 cm)^2/(2 cm)^2
strength2 = 800 kg * 3^2/2^2
strength2 = 800 kg * 9/4
strength2 = 1800 kg
