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Helga [31]
3 years ago
14

Set up the system of equations and then solve it by using an inverse matrix.

Mathematics
1 answer:
ivann1987 [24]3 years ago
8 0

Answer:

x=400000\\y=600000\\z=1000000

Step-by-step explanation:

Given that a trust account manager has $2,000,000 to be invested in three different accounts. The accounts pay 6%, 8%, and 10%, and the goal is to earn $172,000 with the amount invested at 10% equal to the sum of the other two investments.

To accomplish this, assume that x dollars are invested at 8%, y dollars at 10%, and z dollars at 6%.

The equations formed would be

x+y+z= 2000000

Interest amount = 6x+8y+10z =172000(100)

z=x+y\\x+y-z=0

these three can be written in matrix form as

\left[\begin{array}{ccc}1&1&1\\6&8&10\\1&1&-1\end{array}\right] =\left[\begin{array}{ccc}200000\\17200000\\0\end{array}\right]

The inverse of the matrix is

\left[\begin{array}{ccc}9/2&-1/2&-1/2\\-4&1/2&1\\1/2&0&-1/2\end{array}\right]

X = A inverse *B

Using this we get

x=400000\\y=600000\\z=1000000

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The flow velocity measured as in/min at the meter is 5924.13 in/min

<h3 /><h3>Volumetric flow rate</h3>

We know that the volume flow rate Q = Av where

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  • v = flow velocity

Now, Q = 18 gal/min

Converting this to in³/min, we have

Q = 18 gal/min = 18 × 1 gal/min = 18 × 231 in³/min = 4158 in³/min

A = πd²/4 where d = diameter of pipe = 1.0 in.

<h3 /><h3>Flow velocity, v</h3>

Since Q = Av, making v subject of the formula, we have

v = Q/A

v = 4Q/πd²

Substituting the values of the variables into the equation, we have

v = 4Q/πd²

v = 4 × 4158 in³/min ÷ π × (1.0 in)²

v = 16632 in/min ÷ π

v = 5924.13 in/min

So, the flow velocity measured as in/min at the meter is 5924.13 in/min

Learn more about flow velocity here:

brainly.com/question/15648466

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2 years ago
What is 12 times the square root of 16?
Vesna [10]
The answer to this is 48
5 0
3 years ago
(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul
Aloiza [94]

Answer

(a) F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

Step-by-step explanation:

(a) f(t) = 20.5 + 10t + t^2 + δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 f(s)e^{-st} \, dt

where a = ∞

=>  F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt

where d(t) = δ(t)

=> F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt

Integrating, we have:

=> F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3}  )\left \{ {{a} \atop {0}} \right.

Inputting the boundary conditions t = a = ∞, t = 0:

F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}

(b) f(t) = e^{-t} + 4e^{-4t} + te^{-3t}

The Laplace transform of function f(t) is given as:

F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt

F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt

F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt

Integrating, we have:

F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.

Inputting the boundary condition, t = a = ∞, t = 0:

F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}

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THe answer is, 21/5

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Answer:

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Step-by-step explanation:

600m^2 -20m=580m^2

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