Answer:
0.65 is the correct answer
Answer:
9π in², about 28.27 in²
Step-by-step explanation:
The area of a circle is given by the formula ...
A = π·r² . . . . . where r is the radius
Here, the diameter is 12 in, so the radius is 6 in. The area of the entire pizza is then ...
A = π·(6 in)² = 36π in²
Jason ate 1/4 of it, so ate an area that is ...
(1/4)·36π in² = 9π in² ≈ 28.27 in²
_____
<em>Comment on the problem</em>
The problem statement tells you Jason ate 1/4 of a 12-inch pizza. The problem statement does not ask for that quantity to be expressed in any other unit of measure.
Answer:
The values written as Trail mix ingredients is not well written below is the trail mix ingredients
answer :
i) Ian made : 1 3/4 cups of trail mix
ii) Racheal made 1/4 more cup of trail mix
Step-by-step explanation:
<em>Given that :</em>
Trail Mix Ingredients
3
/4 cup almonds
1
/4 cup pumpkin seeds
2
/4 cup coconut
3
/4 cup dried cranberries
1 2
/4 cup walnuts
1 cup granola
2
/4 cup dried bananas
i) Ian used all of coconut , dried cranberries and dried Bananas
= 2/4 + 3/4 + 2/4
Amount of trail mix made = 7/4 = 1 3/4 cups
ii) Racheal made 2 cups of trail mix containing ; all almonds , pumpkin seeds and granola
= 3/4 + 1/4 + 1 = 2 cups of trail mix
Hence Racheal made ( 2 - 1 3/4 ) more trail mix than Ian
= 1/4 more cup of trail mix was made by Racheal
Oh my didn’t answer of why equals 61X equals 2/3 find XY equals 12 on you just multiplied by three and you’ll get the
Check the picture below, so the park looks more or less like so, with the paths in red, so let's find those midpoints.
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad K(\stackrel{x_2}{1}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 1 -3}{2}~~~ ,~~~ \cfrac{ 3 +1}{2} \right) \implies \left(\cfrac{ -2 }{2}~~~ ,~~~ \cfrac{ 4 }{2} \right)\implies JK=(-1~~,~~2) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20J%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20K%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%201%20-3%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%203%20%2B1%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%20-2%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%204%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20JK%3D%28-1~~%2C~~2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ L(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 +5}{2}~~~ ,~~~ \cfrac{ -3 -1}{2} \right) \implies \left(\cfrac{ 4 }{2}~~~ ,~~~ \cfrac{ -4 }{2} \right)\implies LM=(2~~,~~-2) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20L%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-1%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%20-1%20%2B5%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-3%20-1%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%204%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-4%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20LM%3D%282~~%2C~~-2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
now, let's check the other path, JM and KL
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ J(\stackrel{x_1}{-3}~,~\stackrel{y_1}{1})\qquad M(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ -1 -3}{2}~~~ ,~~~ \cfrac{ -3 +1}{2} \right) \implies \left(\cfrac{ -4 }{2}~~~ ,~~~ \cfrac{ -2 }{2} \right)\implies JM=(-2~~,~~-1) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20J%28%5Cstackrel%7Bx_1%7D%7B-3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5Cqquad%20M%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%20-1%20-3%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-3%20%2B1%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%20-4%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-2%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20JM%3D%28-2~~%2C~~-1%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ K(\stackrel{x_1}{1}~,~\stackrel{y_1}{3})\qquad L(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 5 +1}{2}~~~ ,~~~ \cfrac{ -1 +3}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ 2 }{2} \right)\implies KL=(3~~,~~1) \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bmiddle%20point%20of%202%20points%20%7D%20%5C%5C%5C%5C%20K%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B3%7D%29%5Cqquad%20L%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29%20%5Cqquad%20%5Cleft%28%5Ccfrac%7B%20x_2%20%2B%20x_1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20y_2%20%2B%20y_1%7D%7B2%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%5Ccfrac%7B%205%20%2B1%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%20-1%20%2B3%7D%7B2%7D%20%5Cright%29%20%5Cimplies%20%5Cleft%28%5Ccfrac%7B%206%20%7D%7B2%7D~~~%20%2C~~~%20%5Ccfrac%7B%202%20%7D%7B2%7D%20%5Cright%29%5Cimplies%20KL%3D%283~~%2C~~1%29%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
so the red path will be 
