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Andrews [41]
3 years ago
8

Construct the graph of the direct proportion y = kx for k = -1

Mathematics
1 answer:
Sergio [31]3 years ago
8 0

Answer:

In le picture

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What’s the value of this expression?<br><br> 6x4-(5+3) dividend by 2
solong [7]

Answer:

20

Step-by-step explanation:

If you use PEMDAS(Parentheses, Exponents, Multiplication, Division, Addition, Subtraction) you would take care of what's in the parentheses, so you would at 5 and 3 to get 8. Then you would multiply 6 and 4 to get 24, so far your equation looks like this 24 - 8 ÷ 2. You then divide 8 by 2 to get 4 then subtract 24 from 4 to get 20.

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3 years ago
What is the answer for the last 2 questions
arsen [322]
A. is 20ft I believe but I don't know the other one

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3 years ago
Leah likes to stretch 5 minutes for every 10 minutes of dancing. How many minutes should she stretch if she is doing a 50 minute
tangare [24]
Let's set up a proportion.
5/10=?/50
We know that ten times 5 equals 50, so we can multiply the numerator, 5, by 5 to get our answer (what you do to the denominator, you must also do to the numerator).
5x5=25
The answer is 25,
7 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
Please help me it's urgent!
mamaluj [8]

Answer:

consistent system, independent system, independent system, inconsistent system

Step-by-step explanation:

1) A system of linear equation with at least one solution is a consistent system.

2) A system of linear equation with exactly one solution that is called independent system

3) A system of linear equations with an infinite number of solution is a dependent system

4) A system of linear equations with no solution is a inconsistent system

5 0
3 years ago
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