Answer:
<h2>
5,936.76 feet/day</h2>
Step-by-step explanation:
Formula to use to get the speed is expressed as speed = Distance/Time
Given parameters
Distance = 94km
Time = 7.5weeks
Since we are to express the answer in feet per day, we will convert the distance to feet and time to days.
For the distance:
Given the conversion
1 km = 3280.84 feet
95km = (95*3280.84)feet
95km = 311,679.8 feet
For the time:
If 1 week = 7 days
7.5weeks = (7.5 * 7)
7.5weeks = 52.5 days
Speed In ft/day = 311,679.8 feet/ 52.5 days
Speed in ft/day = 5,936.76 feet/day
<em>Hence the speed in feet per day is 5,936.76 feet/day</em>
Answer:is this a real quistion
Step-by-step explanation:?
so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.
in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.
now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B7%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B%5Cfrac%7B19%7D%7B2%7D%7D~%2C~%5Cstackrel%7By_2%7D%7B%5Cfrac%7B7%7D%7B2%7D%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B19%7D%7B2%7D-7%20%5Cright%29%5E2%2B%5Cleft%28%20%5Cfrac%7B7%7D%7B2%7D-4%20%5Cright%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20AM%3D%5Csqrt%7B%5Cleft%28%20%5Cfrac%7B5%7D%7B2%7D%5Cright%29%5E2%2B%5Cleft%28%20-%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Cboxed%7BAM%5Capprox%202.549509756796392%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
The houses can be placed in 362,880 ways.
<u>Step-by-step explanation:</u>
The 9 houses are each in different design.
The each lot can place any of the 9 houses.
- The 1st lot can place anyone house of all the 9 houses.
- The 2nd lot can place one of remaining 8 houses.
- The 3rd lot can place one of remaining 7 houses.
Similarly, the process gets repeated until the last house is placed on a lot.
<u>From the above steps, it can be determined that :</u>
The number of ways to place the 9 houses in 9 lots = 9!
⇒ 9×8×7×6×5×4×3×2×1
⇒ 362880 ways.
Therefore, the houses can be placed in 362880 ways.
Answer:
Dilations produce similar figures because the image and pre-image will have congruent corresponding angles. The corresponding side lengths of the figures will be proportional based on the scale factor. The shape is preserved and the sides are enlarged or reduced by the scale factor
Step-by-step explanation:
