ms sue im sorry about my previous name and i need to know the answer to this, If a firefighter found stray cat stuck in a tree hole, and the circumference of the hole is 3.8 feet and the volume of the cat is 15 cubic inches, would the cat be able to get out if the hole shrunk 19.5 inches? please help i dont understand this and im getting a headache
please help me please if i answer this right i get freee credits in a video game
This is a homework site. Have you finished your school work for the day?
is the cat a cube, or a sphere, or a cylinder, or a disc?
Just knowing its volume does not enable us to compare its cross-section to the size of the hole.
I'm sorry to hear that you're feeling confused, but I'm here to help you out! Let's break down the problem step by step.
First, let's look at the size of the hole. The given information tells us that the circumference of the hole is 3.8 feet. To determine if the cat can get out of the hole, we need to consider whether the cat can fit through the hole after it shrinks.
Secondly, the problem also provides the volume of the cat, which is given as 15 cubic inches. However, this information seems unrelated to the rest of the problem, so we can ignore it for now.
Now let's focus on the shrinking of the hole by 19.5 inches. It's important to note that the hole's circumference is given in feet, while the shrinkage is given in inches. To compare these two measurements, we need to convert them to the same unit. Let's convert the circumference from feet to inches.
The circumference of a circle can be calculated using the formula: C = πd, where C is the circumference and d is the diameter. Since we're given the circumference (C), we can rearrange the formula to solve for the diameter (d): d = C/π.
To convert the 3.8 feet circumference to inches, we need to multiply it by 12 since there are 12 inches in a foot:
3.8 feet * 12 inches/foot = 45.6 inches (approx.)
Now that we have the diameter, we can calculate the new circumference after the hole shrinks by 19.5 inches. Subtracting 19.5 inches from the diameter gives us:
45.6 inches - 19.5 inches = 26.1 inches (approx.)
If the cat's volume is indeed unrelated to the problem, we can conclude that the cat's ability to get out of the hole depends solely on the size of the hole. Since the cat needs to fit through the hole, and the new circumference after shrinking (26.1 inches) is smaller than the previous circumference (45.6 inches), it is likely that the cat will be able to get out of the hole after it shrinks by 19.5 inches.
Remember, it's always important to consider the given information, apply relevant formulas or concepts, and use logical reasoning to arrive at an answer. I hope this explanation helps you understand the problem better!