Answer:
The area of the shaded region is 
Step-by-step explanation:
we know that
The area of the shaded region is equal to the area of the sector of circle of angle 68.9 degrees minus the area of the isosceles triangle
step 1
Find the area of sector of the circle
The area of circle is equal to
assume
substitute
Remember that the area of a circle subtends a central angle of 360 degrees
so
using proportion Find out the area of a sector with a central angle of 68.90 degrees
Let
x -----> the area of a sector
step 2
Find the area of the isosceles triangle
Applying the law of sines
The area is equal to
we have
substitute
step 3
Find the area of the shaded region
Round to the nearest tenth
Well, he only needs 5 gallons of beverages. Because he buys two containers of each, however, that means he bought 320 gallons total (I would say). How I got my answer:
There are 8 pints in a gallon (8 x 2)
There are 4 quarts in a gallon (4 x 2)
There are 16 cups in a gallon (16 x 2)
There are 128 ounces in a gallon (128 x 2)
With that being said, we can add these to find the total amount of gallons of beverages Julius bought. 16 + 8 + 8 + 32 + 256 = 320.
Hopefully this helps!
60cm is the length from A and B
Answer:
Step-by-step explanation: I know that 72/80 is greater than 7/8 because when we make 7/8 as the same denominator as of 72/80 it is only 70/80 which is smaller than 72/80.
the idea behind the recurring decimal as a fraction, is to first off, multiply or divide by some power of 10, in order that we leave the recurring decimal to the right of the decimal point.
then we multiply by a power of 10, in order to move the repeating digits to the left of the decimal point, anyhow, let's proceed.
![\bf 0.6\overline{1212}\implies \cfrac{06.\overline{1212}}{10}\implies \cfrac{6+0.\overline{1212}}{10}\qquad \qquad \stackrel{\textit{now let's make}}{x=0.\overline{1212}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{llll} 100\cdot x &=& 12.\overline{1212}\\\\ &&12+0.\overline{1212}\\\\ &&12+x\\\\ 100x&=&12+x\\\\ 99x&=&12\\\\ x&=&\cfrac{12}{99}\implies x = \cfrac{4}{33} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%200.6%5Coverline%7B1212%7D%5Cimplies%20%5Ccfrac%7B06.%5Coverline%7B1212%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2B0.%5Coverline%7B1212%7D%7D%7B10%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Bnow%20let%27s%20make%7D%7D%7Bx%3D0.%5Coverline%7B1212%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20100%5Ccdot%20x%20%26%3D%26%2012.%5Coverline%7B1212%7D%5C%5C%5C%5C%20%26%2612%2B0.%5Coverline%7B1212%7D%5C%5C%5C%5C%20%26%2612%2Bx%5C%5C%5C%5C%20100x%26%3D%2612%2Bx%5C%5C%5C%5C%2099x%26%3D%2612%5C%5C%5C%5C%20x%26%3D%26%5Ccfrac%7B12%7D%7B99%7D%5Cimplies%20x%20%3D%20%5Ccfrac%7B4%7D%7B33%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \cfrac{06.\overline{1212}}{10}\implies \cfrac{6+x}{10}\implies \cfrac{6+\frac{4}{33}}{10}\implies \cfrac{~~\frac{202}{33}~~}{10}\implies \cfrac{~~\frac{202}{33}~~}{\frac{10}{1}}\implies \cfrac{202}{330} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{101}{165}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B06.%5Coverline%7B1212%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2Bx%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B6%2B%5Cfrac%7B4%7D%7B33%7D%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B202%7D%7B33%7D~~%7D%7B10%7D%5Cimplies%20%5Ccfrac%7B~~%5Cfrac%7B202%7D%7B33%7D~~%7D%7B%5Cfrac%7B10%7D%7B1%7D%7D%5Cimplies%20%5Ccfrac%7B202%7D%7B330%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%5Ccfrac%7B101%7D%7B165%7D~%5Chfill)
notice, we first divided by 10, to move the decimal point over to the right by 1 slot, then we multiplied by 100, to move it two digits over the decimal point, namely the repeating "12", thus we use 100.
