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Vsevolod [243]
2 years ago
5

En el mes de abril en la cancha Toto Hernández de Cucuta, habrán encuentros de

Mathematics
1 answer:
ASHA 777 [7]2 years ago
5 0
Brian’s and the other side is not a big problem with it all I think it is mommy Angela Angela leaving the bath so
You might be interested in
a chef is going to use a mixture of two brands of Italian dressing. The first brand contains 5% vinegar and the second Brian con
tankabanditka [31]

Answer:

  • 273 mL of 5%
  • 117 mL of 15%

Step-by-step explanation:

Let q represent the quantity of 15% dressing used. Then the amount of 5% dressing is (390 -q). The amount of vinegar in the mix is ...

  0.15q + 0.05(390 -q) = 0.08(390)

  0.10q = 31.2 -19.5 = 11.7 . . . . . . subtract 0.05(390) and simplify

  q = 117 . . . . . . . . . . . . . . . . . . multiply by 10

  390-q = 273

The chef should use 273 mL of the first brand (5% vinegar) and 117 mL of the second brand (15% vinegar).

__

<em>Additional comment</em>

You may have noticed that the value of q is (0.08 -0.05)/(0.10 -0.05)×390. The fraction of the mix that is the highest contributor is the ratio of the difference between the mix value and least contributor, divided by the difference between the contributors: (8-5)/(15-5) = 3/10, the fraction that is 15% vinegar. This is the generic solution to mixture problems.

7 0
3 years ago
For lunch, Kile can eat a sandwich with either ham or a bologna and with or without cheese. Kile also has the choice of drinking
Scorpion4ik [409]

Answer:

8

Step-by-step explanation:

Ham with or without cheese-2 choices

Bologna with or without cheese-2 choices

Bologna with cheese with water or juice-2 choices

Bologna without cheese with juice or water-2 choices

Ham with cheese with juice or water -2 choices

Ham without cheese with juice or water -2 choices

2+2+2+2=8

Kile has 8 choices for lunch

4 0
3 years ago
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
Which expression is equivalent to (-2)(a+6)
Eva8 [605]
The equivalent answer to this expression is found by using the distributive property.
(-2a - 12)
Simply multiply the negative 2 into the second parenthesis.
3 0
3 years ago
What is the determinant of <br><br>15<br><br>18<br><br>154
IgorC [24]

Answer:

The determinant is 15.

Step-by-step explanation:

You need to calculate the determinant of the given matrix.

1. Subtract column 3 multiplied by 3 from column 1 (C1=C1−(3)C3):

\left[\begin{array}{ccc}-25&-23&9\\0&3&1\\-5&5&3\end{array}\right]

2. Subtract column 3 multiplied by 3 from column 2 (C2=C2−(3)C3):

\left[\begin{array}{ccc}-25&-23&9\\0&0&1\\-5&-4&3\end{array}\right]

3. Expand along the row 2: (See attached picture).

We get that the answer is 15. The determinant is 15.

6 0
3 years ago
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