The domain:The number of which the logarithm is taken must be greater than 0.
The base of the logarithm must be greater than 0 and not equal to 1.
* greater than 0:
*not equal to 1:
Sum up all the domain restrictions:
The solution:
Now if the base of the logarithm is less than 1, then you need to flip the sign when solving the inequality. If it's greater than 1, the sign remains the same.
* if the base is less than 1:
The inequality:
* if the base is greater than 1:
The inequality:
![\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{the sign remains the same} \\ 2x-2 \leq 1 \\ 2x \leq 3 \\ x \leq \frac{3}{2} \\ x \leq 1 \frac{1}{2} \\ x \in (-\infty, 1 \frac{1}{2}] \\ \\ \hbox{including the condition that the base is greater than 1:} \\ x \in (-\infty, 1 \frac{1}{2}] \ \land \ x \in (2, \infty) \\ \Downarrow \\ x \in \emptyset](https://tex.z-dn.net/?f=%5Clog_%7B8x%5E2-23x%2B15%7D%20%282x-2%29%20%5Cleq%20%5Clog_%7B8x%5E2-23x%2B15%7D%201%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%7C%5Chbox%7Bthe%20sign%20remains%20the%20same%7D%20%5C%5C%202x-2%20%5Cleq%201%20%5C%5C%202x%20%5Cleq%203%20%5C%5C%20x%20%5Cleq%20%5Cfrac%7B3%7D%7B2%7D%20%5C%5C%20x%20%5Cleq%201%20%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%20x%20%5Cin%20%28-%5Cinfty%2C%201%20%5Cfrac%7B1%7D%7B2%7D%5D%20%5C%5C%20%5C%5C%20%5Chbox%7Bincluding%20the%20condition%20that%20the%20base%20is%20greater%20than%201%3A%7D%20%5C%5C%20x%20%5Cin%20%28-%5Cinfty%2C%201%20%5Cfrac%7B1%7D%7B2%7D%5D%20%5C%20%5Cland%20%5C%20x%20%5Cin%20%282%2C%20%5Cinfty%29%20%5C%5C%20%5CDownarrow%20%5C%5C%20x%20%5Cin%20%5Cemptyset)
Sum up both solutions:
The final answer is:
Answer:
32π is the volume of the larger sphere
Step-by-step explanation:
The volume of a sphere is V = 4/3πr³. The smaller sphere is V = 4π. This means r³ divided by 3 was 1. So r³ = 3. For the larger sphere, substitute r = 2r into the volume formula.
V = 4/3πr³
V = 4/3π(2r)³
V = 4/3π8r³
We know from the previous problem that r³ = 3.
So the volume formula simplifies to V = 4π(8) = 32π
Complete question is;
Many states run lotteries to raise money. A website advertises that it knows "how to increase YOUR chances of Winning the Lottery." They offer several systems and criticize others as foolish. One system is called Lucky Numbers. People who play the Lucky Numbers system just pick a "lucky" number to play, but maybe some numbers are luckier than others. Let's use a simulation to see how well this system works. To make the situation manageable, simulate a simple lottery in which a single digit from 0 to 9 is selected as the winning number. Any value can be picked, but for this exercise, pick 1 as the lucky number. What proportion of the time do you win?
Answer:
10%
Step-by-step explanation:
We are told that To make the situation manageable, simulate a simple lottery in which a single digit from 0 to 9 is selected as the winning number.
This means the total number of single digits that could possibly be a winning one is 10.
Since we are told that only 1 can be picked, thus;
Probability of winning is; 1/10 = 0.1 or 10%
Answer:
1. fraction is 64/100 and decimal is 0.64
2. 12.5 is percent, decimal is 0.125
3. 140 is percent, I think is 1 2/5
4. 2.75 is decimal, 275 as percent
5. 0.08 is decimal, I think is 16/25
