<span>1, 7, 23, 161 is what can go into 161</span>
Answer:
48 mph
Step-by-step explanation:
First we need to find the distance from Elkhart to Chicago. Toledo to Elkhart is 136 miles and Toledo to Chicago 244 miles.
So the distance from Elkhart to Chicago can be calculated, since Chicago is farther from Toledo than Elkhart, as: distance(Toledo to Chicago) - distance(Toledo to Elkhart). These distances are given in the problem, so the distance from Elkhat to Chicago is: 244 miles - 136 miles = 108 miles.
This problem basically wants to know the slowest you can be yet still ariving on time. If you are the minimum speed, you will arrive in Chicago exactly at 10:30 A.M. So you have 2 hours and 15 minutes(10:30 A.M - 8.15A.M.) to drive 108 miles.
15 minutes is a fourth of a hour, so you have 2.25hours to go through 108 miles.
The minimum speed you must maintain is 108mph/2.25h = 48mph.
The equation would be ' y = 1/2x'
The photo of my working will be attached.
Answer:
X=30
Step-by-step explanation:
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Answer:
The volume of the cone is approximately 453.0 cm³
Step-by-step explanation:
The volume of a cone is one third that of a cylinder with the same height and radius. That gives us 1/3 πr²h, where r is radius and h is height.
However, we are not given the height of the cone, but the side length. We can work out the height using the Pythagorean theorem, as we have a right triangle with the height, base radius, and length. You may recall that the Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of it's other two sides:
![a^2 = b^2 + c^2](https://tex.z-dn.net/?f=a%5E2%20%3D%20b%5E2%20%2B%20c%5E2)
So we can find the height of the cone with that:
![10.8^2 = 7.4^2 + h^2\\h^2 = 10.8^2 - 7.4^2\\h^2 = 116.64 - 54.76\\h^2 = 61.88\\h = \sqrt{61.88}\\h \approx 7.9](https://tex.z-dn.net/?f=10.8%5E2%20%3D%207.4%5E2%20%2B%20h%5E2%5C%5Ch%5E2%20%3D%2010.8%5E2%20-%207.4%5E2%5C%5Ch%5E2%20%3D%20116.64%20-%2054.76%5C%5Ch%5E2%20%3D%2061.88%5C%5Ch%20%3D%20%5Csqrt%7B61.88%7D%5C%5Ch%20%5Capprox%20%207.9)
Now that we have the cone's height, we can solve for its volume:
![v = \frac{1}{3} \pi r^2 h\\v = \frac{1}{3} \pi \times 7.4^2 \times 7.9\\v = \frac{1}{3} \pi \times 54.76 \times 7.9\\v = \frac{1}{3} \pi \times 432.6\\v = \pi \times 144.2\\v \approx 453.0 cm^3](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20r%5E2%20h%5C%5Cv%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20%5Ctimes%207.4%5E2%20%20%5Ctimes%207.9%5C%5Cv%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20%5Ctimes%2054.76%20%5Ctimes%207.9%5C%5Cv%20%3D%20%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%20%5Ctimes%20432.6%5C%5Cv%20%3D%20%5Cpi%20%5Ctimes%20144.2%5C%5Cv%20%5Capprox%20453.0%20cm%5E3)