What is the simplified version of -3/5
2 answers:
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The question is incomplete. The map is attached as a photo and here is the complete question:
a. What is the actual distance between Saugerties and
Kingston? ___________________________
b. Catskill is 15 miles from Saugerties. What would the
distance on the map be? ___________________________
c. On another map, the distance between Saugerties and
Kingston is 2 inches. What would the distance from Saugerties to
Catskill be on this map? __________________
Since you have asked the answer for part (c) only, here is the answer:
Answer:
<u>The distance from Saugerties to Catskill would be </u><u>3 inches</u><u> on this map.</u>
Step-by-step explanation:
On the given map, the distance between Saugerties and Kingston is 4 inches and the scale is <u>1 inch = 2.5 miles.</u>
To calculate the actual distance in miles between Saugerties and Kingston, consider the given scale:
1 inch ---------- 2.5 miles
4 inches ------ x miles
Cross multiplying:
1x = 2.5 x 4
x = 10 miles
The actual distance between Saugerties and Kingston is 10 miles.
On another map, the distance between Saugerties and Kingston is 2 inches. Which means the scale is 2 inches = 10 miles. So 1 inch = 10/2 = 5 miles
The scale on the other map is 1 inch = 5 miles.
Catskill is 15 miles from Saugerties (given in the previous part). So on the map:
1 inch -------- 5 miles
y inch ------- 15 miles
Cross multiplying:
15 = 5y
y = 15/5
<u>y = 3 inches.</u>
The distance from Saugerties to Catskill would be 3 inches on this map.
Answer:
Step-by-step explanation:
We know that the length is four times the width, so:
We also know the area, which is 324 m². The formula for area:
Insert the known values:
Solve for w. Simplify by removing parentheses:
Divide 4 from both sides to isolate the variable:
Find the square root of both sides:
The width is 9 m.
We know the width. Now find the length by using the area formula and inserting known values:
Solve for l. Divide both sides by 9:
The length of the rectangle is 36. (You can check: 4 times 9 is 36)
Now find the perimeter:
Insert values:
The perimeter is 90 m.
Check the picture below.
so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -2 &,& -4~) % (c,d) &&(~ 3 &,& 8~) \end{array}~~~ % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[3-(-2)]^2+[8-(-4)]^2}\implies d=\sqrt{(3+2)^2+(8+4)^2} \\\\\\ d=\sqrt{25+144}\implies d=\sqrt{169}\implies d=13\qquad\qquad \qquad \stackrel{radius}{\frac{13}{2}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%20-2%20%26%2C%26%20-4~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%203%20%26%2C%26%208~%29%0A%5Cend%7Barray%7D~~~%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B3-%28-2%29%5D%5E2%2B%5B8-%28-4%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%283%2B2%29%5E2%2B%288%2B4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B25%2B144%7D%5Cimplies%20d%3D%5Csqrt%7B169%7D%5Cimplies%20d%3D13%5Cqquad%5Cqquad%20%5Cqquad%20%20%5Cstackrel%7Bradius%7D%7B%5Cfrac%7B13%7D%7B2%7D%7D)
so, the center of the circle is the midpoint of that diametrical segment, and half that length is the radius.