If the amount Damien deposits into the Bank of Chance is x, then the amount he deposits into the Merchant Bank is 3*x (as he deposits 3 times as much there). Since 3*x+x=4x, and we know that Utopian Financing's deposit is 20% (or 0.2 by moving the decimal in 20.00 2 places over to the left), 4x*0.2=0.8x=amount he invests in Utopian Financing. Adding them all up, we get x+3x+0.8x=1200=4.8x. Dividing both sides by 4.8, we get 1200/4.8=x=
250 dollars invested in the Bank of Chance. To find 0.8 of 250, we multiply the two to get 0.8*250=200=amount invested into Utopian Financing
Answer:
x = 41°, y = 139°
Step-by-step explanation:
The given parameters are;
Line <em>m</em> is parallel to line <em>n</em> and lines <em>m</em> and <em>n</em> have a common transversal
The corresponding angles formed by the common transversal to the two parallel lines are 41° on line <em>m</em> and <em>x°</em> on line <em>n</em>
Therefore, x° = 41° by corresponding angles formed between on two parallel lines by a common transversal are equal
x° and y° are linear pair angles and they are, supplementary
∴ x° + y° = 180°
∴ x° + y° = 41° + y° = 180°
y° = 180° - 41° = 139°
y° = 139°.
Answer:
a = 20; b = 2√101
Step-by-step explanation:
The triangle at upper left and the one at lower right are similar. Ratios of corresponding sides are equal, so ...
200/a = a/2
400 = a^2
√400 = 20 = a
Then, by the Pythagorean theorem, the value of b can be found:
b = √(a^2 + 2^2) = √(400 +4) = √4·√101
b = 2√101
recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?
![\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bccll%7D%20miles%26hours%5C%5C%20%5Ccline%7B1-2%7D%2012%261%5C%5C%2018%26x%20%5Cend%7Barray%7D%5Cimplies%20%5Ccfrac%7B12%7D%7B18%7D%3D%5Ccfrac%7B1%7D%7Bx%7D%5Cimplies%2012x%3D18%5Cimplies%20x%3D%5Ccfrac%7B18%7D%7B12%7D%5Cimplies%20x%3D%5Ccfrac%7B3%7D%7B2%7D)
so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)
![\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20%5Cunderline%7Bd%7D%20in%20the%202nd%20equation%7D%7D%7B%5Cboxed%7B%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%7D%3D16t%7D%5Cimplies%2012t%2B18%3D16t%20%5C%5C%5C%5C%5C%5C%2018%3D4t%5Cimplies%20%5Ccfrac%7B18%7D%7B4%7D%3Dt%5Cimplies%20%5Ccfrac%7B9%7D%7B2%7D%3Dt%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bfour%20and%20a%20half%20hours%7D%7D%7B4%5Cfrac%7B1%7D%7B2%7D%3Dt%7D)