24. armin is trying to decide whether to buy a season pass to his college basketball teams 20 home games…
24. armin is trying to decide whether to buy a season pass to his college basketball teams 20 home games this season. the cost of an individual ticket is $14, and the cost of a season pass is $175. the season pass will admit armin to any home basketball game at no additional cost. what is the minimum number of home basketball games armin must attend this season in order for the cost of a season pass to be less than the total cost of buying an individual ticket for each game he attends? f. 8 g. 9 h. 12 j. 13 k. 20 25. (4.8×10^(-7))/(1.6×10^(-11)) =? a. 3.0×10^4 b. 3.0×10^(-4) c. 3.0×10^(-18) d. 3.2×10^18 e. 3.2×10^4 26. a circle in the standard (x,y) coordinate plane has center c(-1,2) and passes through a(2,6). line segment ab is a diameter of this circle. what are the coordinates of point b? f. (-6,-2) g. (-5,-1) h. (-4,-2) j. (4,2) k. (5,10) 27. which of the following expressions is a factor of x^3 - 64? a. x - 4 b. x + 4 c. x + 64 d. x^2+16 e. x^2 - 4x + 16
Answer
24.
Explanation:
Step1: Set up inequality
Let $n$ be the number of games. We want $175<14n$.
Step2: Solve for $n$
$n > \frac{175}{14}=12.5$. Since $n$ must be an integer, the minimum value of $n$ is $13$.
Answer:
J. 13
25.
Explanation:
Step1: Divide the coefficients
$\frac{4.8}{1.6}=3$.
Step2: Use exponent - rule for division
$\frac{10^{-7}}{10^{-11}}=10^{-7-(-11)} = 10^{4}$.
Step3: Combine results
The result is $3\times10^{4}$.
Answer:
A. $3.0\times10^{4}$
26.
Explanation:
Step1: Recall mid - point formula
The mid - point of a diameter with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is the center of the circle. The mid - point formula is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. Let the center $C(-1,2)$ be the mid - point of $\overline{AB}$ where $A(2,6)$ and $B(x,y)$.
Step2: Solve for $x$
$\frac{2 + x}{2}=-1$, then $2+x=-2$, so $x=-4$.
Step3: Solve for $y$
$\frac{6 + y}{2}=2$, then $6 + y = 4$, so $y=-2$.
Answer:
H. $(-4,-2)$
27.
Explanation:
Step1: Use the difference of cubes formula
$a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})$. Here $a=x$ and $b = 4$ since $x^{3}-64=x^{3}-4^{3}=(x - 4)(x^{2}+4x + 16)$.
Answer:
A. $x - 4$