Question

1. a rock is thrown into a lake causing ripples to form in concentric circles. the radius (in feet) of the outermost ripple is r(t)=0.8t, where t is the number of seconds after the rock hits the water. the area of the circle is given by the function a(r)=πr². a. find (a∘r)(t) and explain what information can be determined using this composition. b. what is the area of the outermost ring 6 seconds after the rock hits the water? 6. your teacher offers to give the whole class a bonus if everyone passes the next math test. the teacher says she will give everyone a 10 - point bonus and increase everyones grade by 9%. let s represent a students initial score on the test. a. write an equation, t(s), that represents a students new score if they were to receive just the 10 - point bonus. b. write an equation, n(s), that represents a students new score if they were to receive just the 9% bonus. c. write a composition to model a students new score if they receive the 10 - point bump, then receive the 9% increase. d. write a composition to model receiving the 9% increase, then the 10 - point bump. e. if you score a 75 on the test, which option (c or d) would you prefer? provide calculations to support your conclusion. 7. your favorite restaurant is offering a 15% discount on their entree menu items. you have a coupon valued at $5.00. a. write a function to model the discount and to model applying the coupon (make sure to clearly define any variables used, as well to identify which equation is which). b. write a composition to model the cost of an entree if the discount is applied, then the coupon. c. write a composition to model applying the coupon followed by the discount. d. which is most cost effective? provide calculations to support your conclusion.

Explanation:

5.

Step1: Find the composition $(A\circ r)(t)$

The composition of functions $(A\circ r)(t)=A(r(t))$. Given $r(t) = 0.8t$ and $A(r)=\pi r^{2}$, we substitute $r = 0.8t$ into $A(r)$. So, $A(r(t))=\pi(0.8t)^{2}=0.64\pi t^{2}$. This composition gives the area of the out - most ripple as a function of time $t$ (in seconds) after the rock hits the water.

Step2: Find the area at $t = 6$

Substitute $t = 6$ into $(A\circ r)(t)$. $(A\circ r)(6)=0.64\pi(6)^{2}=0.64\pi\times36 = 23.04\pi\approx 72.38$ square feet.

Step1: Equation for 10 - point bonus

If $s$ is the initial score, then $T(s)=s + 10$.

Step2: Equation for 9% bonus

If $s$ is the initial score, then $N(s)=s+0.09s=1.09s$.

Step3: Composition for 10 - point then 9% bonus

$(N\circ T)(s)=N(T(s))$. Since $T(s)=s + 10$, then $N(T(s))=1.09(s + 10)=1.09s+10.9$.

Step4: Composition for 9% then 10 - point bonus

$(T\circ N)(s)=T(N(s))$. Since $N(s)=1.09s$, then $T(N(s))=1.09s + 10$.

Step5: Compare for $s = 75$

For $(N\circ T)(75)=1.09\times75+10.9=81.75+10.9 = 92.65$.
For $(T\circ N)(75)=1.09\times75+10=81.75 + 10=91.75$.

Step1: Define functions for discount and coupon

Let $x$ be the original price of an entree. The function for the 15% discount is $D(x)=(1 - 0.15)x=0.85x$. The function for the $5 - dollar$ coupon is $C(x)=x - 5$.

Step2: Composition for discount then coupon

$(C\circ D)(x)=C(D(x))$. Since $D(x)=0.85x$, then $C(D(x))=0.85x-5$.

Step3: Composition for coupon then discount

$(D\circ C)(x)=D(C(x))$. Since $C(x)=x - 5$, then $D(C(x))=0.85(x - 5)=0.85x-4.25$.

Step4: Compare for cost - effectiveness

Let's assume the original price $x$ of an entree.
$(C\circ D)(x)- (D\circ C)(x)=(0.85x-5)-(0.85x - 4.25)=- 0.75$. So, applying the discount first and then the coupon is more cost - effective.

Answer:

a. $(A\circ r)(t)=0.64\pi t^{2}$, it gives the area of the out - most ripple as a function of time $t$ after the rock hits the water.
b. Approximately $72.38$ square feet

6.