Answer:
A = 0.8 litres
B = 0.7 litres
C = 0.5 litres
D = 0.2 litres
Step-by-step explanation
Here's what we know:
1. Jug A = B + .1 litres
2. Jug C = B - 200 (or 0.2 litres)
3. Jug D = .25 x A
4. Jug A + Jug B = 1.5 litres
In problem 1, we learned that Jug A has .1 litres more than Jug B and in problem 4, the two of them added together are 1.5 litres. To solve this we can combine the problems.
B + .1 litres + B = 1.5 litres
2B + .1 = 1.5
Subtract .01 from each side and you have 2B = 1.4
Divide each side by 2 and you have B = 0.7 litres
Plug this info into problem 1 and you can solve for A. (0.7 + 0.1 = 0.8)
Plug this info into problem 2 and you can solve for C. (0.7 - 0.2 = 0.5)
Since you have A, you can use that info to solve problem 3 (0.25 x 0.8 = 0.2)
Answer is in the file below
tinyurl.com/wpazsebu
Answer:
188.4
Step-by-step explanation:
because if you multiply 15.7 and 12 you get 188.4 and that is the area of the jumping space.
2×4=8 because 2 and 4 would be the addends and their less than 5 and it equal an answer between 4 and 10
well, let's start off by doing some grouping, what we'll be doing is so-called "completing the square" as in a perfect square trinomial, since that's what the vertex form of a quadratic uses.
![\bf f(x) = (x^2+6x)+14\implies f(x) = (x^2+6x+\boxed{?}^2)+14](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%20%3D%20%28x%5E2%2B6x%29%2B14%5Cimplies%20f%28x%29%20%3D%20%28x%5E2%2B6x%2B%5Cboxed%7B%3F%7D%5E2%29%2B14)
well, darn, we have a missing number for our perfect trinomial, however let's recall that in a perfect square trinomial the middle term is really the product of 2 times the term on the left and the term on the right without the exponent, so then we know that
![\bf 2(x)\boxed{?} = 6x\implies \boxed{?}=\cfrac{6x}{2x}\implies \boxed{?}=3](https://tex.z-dn.net/?f=%5Cbf%202%28x%29%5Cboxed%7B%3F%7D%20%3D%206x%5Cimplies%20%5Cboxed%7B%3F%7D%3D%5Ccfrac%7B6x%7D%7B2x%7D%5Cimplies%20%5Cboxed%7B%3F%7D%3D3)
well then, that's our mystery guy, now, let's recall all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add 3², we also have to subtract 3².
![\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ f(x) = (x^2+6x)+14\implies f(x) = (x^2+6x+3^2-3^2)+14 \\\\\\ f(x) = (x^2+6x+3^2)+14-3^2\implies f(x) = (x+3)^2+5~\hfill \stackrel{vertex}{(-3,5)}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bvertical%20parabola%20vertex%20form%7D%20%5C%5C%5C%5C%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cbegin%7Bcases%7D%20%5Cstackrel%7Bvertex%7D%7B%28h%2Ck%29%7D%5C%5C%5C%5C%20%5Cstackrel%7B%22a%22~is~negative%7D%7Bop%20ens~%5Ccap%7D%5Cqquad%20%5Cstackrel%7B%22a%22~is~positive%7D%7Bop%20ens~%5Ccup%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20f%28x%29%20%3D%20%28x%5E2%2B6x%29%2B14%5Cimplies%20f%28x%29%20%3D%20%28x%5E2%2B6x%2B3%5E2-3%5E2%29%2B14%20%5C%5C%5C%5C%5C%5C%20f%28x%29%20%3D%20%28x%5E2%2B6x%2B3%5E2%29%2B14-3%5E2%5Cimplies%20f%28x%29%20%3D%20%28x%2B3%29%5E2%2B5~%5Chfill%20%5Cstackrel%7Bvertex%7D%7B%28-3%2C5%29%7D)