Answer:
3048 x 1.04 ^3 = 3428.585474
Answer:
y=-5/3x+20
Step-by-step explanation:
Let the equation of the required line be represented as ![\[y=mx+c\]](https://tex.z-dn.net/?f=%5C%5By%3Dmx%2Bc%5C%5D)
This line is perpendicular to the line ![\[y=\frac{3}{5}x+10\]](https://tex.z-dn.net/?f=%5C%5By%3D%5Cfrac%7B3%7D%7B5%7Dx%2B10%5C%5D)
![\[=>m*\frac{3}{5}=-1\]](https://tex.z-dn.net/?f=%5C%5B%3D%3Em%2A%5Cfrac%7B3%7D%7B5%7D%3D-1%5C%5D)
![\[=>m=\frac{-5}{3}\]](https://tex.z-dn.net/?f=%5C%5B%3D%3Em%3D%5Cfrac%7B-5%7D%7B3%7D%5C%5D)
So the equation of the required line becomes ![\[y=\frac{-5}{3}x+c\]](https://tex.z-dn.net/?f=%5C%5By%3D%5Cfrac%7B-5%7D%7B3%7Dx%2Bc%5C%5D)
This line passes through the point (15.-5)
![\[-5=\frac{-5}{3}*15+c\]](https://tex.z-dn.net/?f=%5C%5B-5%3D%5Cfrac%7B-5%7D%7B3%7D%2A15%2Bc%5C%5D)
![\[=>c=20\]](https://tex.z-dn.net/?f=%5C%5B%3D%3Ec%3D20%5C%5D)
So the equation of the required line is ![\[y=\frac{-5}{3}x+20\]](https://tex.z-dn.net/?f=%5C%5By%3D%5Cfrac%7B-5%7D%7B3%7Dx%2B20%5C%5D)
Among the given options, option 4 is the correct one.
Answer:
10
Step-by-step explanation:
because of the math envolved
Answer:
1/2 glass milk is left.
Step-by-step explanation:
2/3 - 1/6 = 1/2
Answer:
A) 150 m
B) 180.28 m
Step-by-step explanation:
A) In order to find the horizontal distance from the base of the cliff to the speed boat, use the Pythagorean theorem to calculate the length of the missing side.
We are told that one of the sides is 80 m and the hypotenuse is 170 m. Therefore,
- a² + b² = c²
- (80)² + b² = (170)²
- b² = 170² - 80²
- b² = 22500
- b = 150
The horizontal distance between the base of the cliff and the boat is 150 m.
B) Now, the side that was 80 m is now 100 m (includes the height that the helicopter is above Jumbo). The horizontal distance remains the same, 150 m, but the hypotenuse is different. Solving for the hypotenuse will give us the distance between the helicopter and the speed boat.
- (100)² + (150)² = c²
- 32500 = c²
- 180.28 = c
The distance between the helicopter and the speed boat is 180.28 m.
