Question

1. scientists modeled the intensity of the sun, $i$, as a function of the number of hours since 6:00 am, $h$, using the function $i(h)=\frac{12h - h^{2}}{36}$. they model the fahrenheit temperature of the soil, $t$, as a function of the intensity using the function $t(i)=sqrt{5000i}$. which of the following is closest to the temperature of the soil at 2:00 pm? (1) 54 (3) 67 (2) 84 (4) 38 7. physics students are studying the effect of the temperature on the speed of sound. they find that the speed of sound in meters per second is a function of the temperature in degrees kelvin, $k$, given by $s(k)=sqrt{410k}$. kelvin is a function of the temperature in celsius given by $k(c)=c + 273.15$. find the speed of sound when the temperature is 30 degrees celsius. round to the nearest tenth.

Explanation:

Step1: Calculate $h$ for 2:00 pm

From 6:00 am to 2:00 pm, $h = 8$.

Step2: Find the intensity $I$ at $h = 8$

Substitute $h = 8$ into $I(h)=\frac{12h - h^{2}}{36}$, so $I(8)=\frac{12\times8 - 8^{2}}{36}=\frac{96 - 64}{36}=\frac{32}{36}=\frac{8}{9}$.

Step3: Calculate the soil - temperature $T$

Substitute $I=\frac{8}{9}$ into $T(I)=\sqrt{5000I}$, so $T(\frac{8}{9})=\sqrt{5000\times\frac{8}{9}}=\sqrt{\frac{40000}{9}}=\frac{200}{3}\approx67$.

for question 7:

Step1: Calculate $K$ when $C = 30$

Substitute $C = 30$ into $K(C)=C + 273.15$, so $K(30)=30 + 273.15=303.15$.

Step2: Calculate the speed of sound $S$

Substitute $K = 303.15$ into $S(K)=\sqrt{410K}$, so $S(303.15)=\sqrt{410\times303.15}=\sqrt{124291.5}\approx352.6$.

Answer:

(3) 67