Answer: the rate at which the distance between the boats is increasing is 68 mph
Step-by-step explanation:
The direction of movement of both boats forms a right angle triangle. The distance travelled due south and due east by both boats represents the legs of the triangle. Their distance apart after t hours represents the hypotenuse of the right angle triangle.
Let x represent the length the shorter leg(south) of the right angle triangle.
Let y represent the length the longer leg(east) of the right angle triangle.
Let z represent the hypotenuse.
Applying Pythagoras theorem
Hypotenuse² = opposite side² + adjacent side²
Therefore
z² = x² + y²
To determine the rate at which the distances are changing, we would differentiate with respect to t. It becomes
2zdz/dt = 2xdx/dt + 2ydy/dt- - - -- - -1
One travels south at 32 mi/h and the other travels east at 60 mi/h. It means that
dx/dt = 32
dy/dt = 60
Distance = speed × time
Since t = 0.5 hour, then
x = 32 × 0.5 = 16 miles
y = 60 × 0.5 = 30 miles
z² = 16² + 30² = 256 + 900
z = √1156
z = 34 miles
Substituting these values into equation 1, it becomes
2 × 34 × dz/dt = (2 × 16 × 32) + 2 × 30 × 60
68dz/dt = 1024 + 3600
68dz/dt = 4624
dz/dt = 4624/68
dz/dt = 68 mph
Answer:
(x - 5)² + (y + 3)² = 16
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
here (h, k) = (5, - 3) and r = 4, thus
(x - 5)² + (y - (- 3))² = 4², that is
(x - 5)² + (y + 3)² = 16 ← equation of circle
Answer:
The lower bound of 19.4 is 19.35
The sequence is geometric, so
![a_n = r a_{n-1}](https://tex.z-dn.net/?f=a_n%20%3D%20r%20a_%7Bn-1%7D)
for some constant r. From this rule, it follows that
![a_3 = r a_2 \implies 20 = 2r \implies r = 10](https://tex.z-dn.net/?f=a_3%20%3D%20r%20a_2%20%5Cimplies%2020%20%3D%202r%20%5Cimplies%20r%20%3D%2010)
and we can determine the first term to be
![a_2 = r a_1 \implies 2 = 10 a_1 \implies a_1 = \dfrac15](https://tex.z-dn.net/?f=a_2%20%3D%20r%20a_1%20%5Cimplies%202%20%3D%2010%20a_1%20%5Cimplies%20a_1%20%3D%20%5Cdfrac15)
Now, by substitution we have
![a_n = r a_{n-1} = r^2 a_{n-2} = r^3 a_{n-3} = \cdots](https://tex.z-dn.net/?f=a_n%20%3D%20r%20a_%7Bn-1%7D%20%3D%20r%5E2%20a_%7Bn-2%7D%20%3D%20r%5E3%20a_%7Bn-3%7D%20%3D%20%5Ccdots)
and so on down to (D)
![a_n = r^{n-1} a_1 = 10^{n-1} \cdot \dfrac15](https://tex.z-dn.net/?f=a_n%20%3D%20r%5E%7Bn-1%7D%20a_1%20%3D%2010%5E%7Bn-1%7D%20%5Ccdot%20%5Cdfrac15)
(notice how the exponent on r and the subscript on a add up to n)
Answer:
The sample space is:
- (T,N): table height and brown
- (T, W): table height and white
- (T, K) : table height and black
- (B, N): bar height and brown
- (B, W): bar height and white
- (B, K): bar height and black
- (X, N): XL height and brown
- (X, W): XL height and white
- (W, K): XL height and black
Explanation:
The <em>sample space </em>is the set of all the possible outputs, i.e. the possible different combinations that can be choosen.
Use the letters T, B, and X to represent, respectively, table height, bar height, and XL height,
Use letters N, W, and K to represent, respectively, the colors brown, white and black.
Each combination consists of a height (T, B or X) and a color (N, W, K); thus, your sample space shall have 3 × 3 different combinations. These are:
- (T,N): table height and brown
- (T, W): table height and white
- (T, K) : table height and black
- (B, N): bar height and brown
- (B, W): bar height and white
- (B, K): bar height and black
- (X, N): XL height and brown
- (X, W): XL height and white
- (W, K): XL height and black