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3 years ago

What is -5over 21 minus -1 over fourtynine

Mathematics

1 answer:

Margaret [11]3 years ago

6 0

The answer to that is -32 over 147. I have a TI-84 plus CE calculator so i can get fractions in it

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Please help !! thank you!

11Alexandr11 [23.1K]

Wow woqlqnw e let me know if I need more help

6 0

3 years ago

Jason bought 8 new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There are n

lyudmila [28]

Answer:

66 cards

Step-by-step explanation:

41 - 8 = 33

33 times 2 = 66

6 0

3 years ago

-x+y+z=1, x-y+z=2, x+y-z=1

Virty [35]

Answer:

Step-by-step explanation:

Given the simultaneous equation

-x+y+z=1... 1

x-y+z=2 ... 2

x+y-z=1 .... 3

Add equation 1 and 2;

-x+x + y - y + z + z = 1+2

0 + 0 + 2z = 3

2z = 3

z = 3/2

Add equation 2 and 3

x+x = 2 + 1

2x = 3

x = 3/2

Substitute x = 3/2 and z = 3/2 into 3

From 3;

x+y-z = 1

3/2 + y - 3/2 = 1

y + 0 = 1

y = 1

<em>Hence the solution (x, y, z) is (3/2, 1, 3/2)</em>

6 0

4 years ago

What's 4.3 rounded to the nearest tenth

kolezko [41]

The answer would be 4.0

4 0

4 years ago

Read 2 more answers

A 50-gal tank initially contains 10 gal of fresh water. At t = 0, a brine solution

scZoUnD [109]

$\huge \mathbb{SOLUTION:}$

$\begin{array}{l} \textsf{Let }A(t)\textsf{ be the function which gives the amount} \\ \textsf{of the salt dissolved in the liquid in the tank at} \\ \textsf{any time }t. \textsf{ We want to develop a differential} \\ \textsf{equation that, when solved, will give us an} \\ \textsf{expression for }A(t). \\ \\ \textsf{The basic principle determining the differential} \\ \textsf{equation is} \\ \\ \end{array}$

$\boxed{ \footnotesize \begin{array}{l} \qquad\quad \quad\Large{\dfrac{dA}{dt} = R_{in} - R_{out}} \\ \\ \textsf{where:} \\ \\ \begin{aligned} \bullet\: R_{in} &= \textsf{rate of the salt entering} \\ &= \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt inflow}\end{array}}\right) \times \small(\textsf{Input of brine}) \\ \\ \bullet\: R_{out} &= \textsf{rate of the salt leaving} \\ &= \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt outflow}\end{array}}\right) \times \small(\textsf{Output of brine}) \end{aligned} \end{array}} \\ \\$

$\begin{array}{l} \textsf{On the problem, the amount of salt in the tank,} \\ A(t), \textsf{changes overtime is given by the differential} \\ \textsf{equation} \\ \\ \footnotesize A'(t) = \left(\dfrac{4\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{1\ \textsf{lb}}{1\ \textsf{gal}}\right) - \left(\dfrac{2\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{A(t)\ \textsf{lb}}{10 + (4 - 2)t\ \textsf{gal}}\right) \\ \\ \textsf{There's no salt in the tank (fresh water) at the} \\ \textsf{start, so }A(0) = 0. \textsf{ The amount of solution in the} \\ \textsf{tank is given by }10 + (4 -2)t, \textsf{so the tank will} \\ \textsf{overflow once this expression is equal to the total} \\ \textsf{volume or capacity of the tank.} \\ \\ 10 + (4 - 2)t = 50 \\ \\ \textsf{Solving for }t,\textsf{ we get} \\ \\ \implies \boxed{t = 20\textsf{ mins}} \\ \\ A'(t) = 4 - \dfrac{2A(t)}{10 + 2t} \\ \\ A'(t) = 4 - \dfrac{1}{5 + t} A(t) \\ \\ A'(t) + \dfrac{1}{5 + t} A(t) = 4 \\ \\ \textsf{This is a linear ODE with integrating factor} \\ \mu (t) = e^{\int \frac{1}{5 + t}\ dt} = e^{\ln |5 + t|} = 5 + t \\ \\ \textsf{Multiplying this to the ODE, we get} \\ \\ (5 + t)A'(t) + A(t) = 4(5 + t) \\ \\ [(5 + t)A(t)]' = 20 + 4t \\ \\ (5 + t)A(t) = 20t + 2t^2 + C \\ \\ \textsf{Since }A(0) = 0, \textsf{ we get } C = 0. \\ \\ A(t) = \dfrac{2t^2 + 20t}{t + 5} \\ \\ A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{So the function that gives the amount of salt at} \\ \textsf{any given time }t,\textsf{ is given by} \\ \\ \implies A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{The amount of salt in the tank at the moment} \\ \textsf{of overflow or at }t = 20\textsf{ mins is equal to} \\ \\ A(20) = 2(20) + 10 - \dfrac{50}{20 + 5} \\ \\ \implies \boxed{A = 48\ \textsf{gallons}} \end{array}$

$\Large \mathbb{ANSWER:}$

$\qquad\red{\boxed{\begin{array}{l} \textsf{a. }20\textsf{ mins} \\ \\ \textsf{b. }48\textsf{ gallons}\end{array}}}$

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5 0

3 years ago