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Anna71 [15]
3 years ago
11

Which steps can be used in order to determine the solution to -1.3+4.6x=0.3+4x?

Mathematics
2 answers:
SpyIntel [72]3 years ago
8 0
-1.3+4.6x=0.3+4x
-0.3. -0.3
-1.6+4.6x=4x
-4.6x. -4.6x
-1.6=-0.6x
___. ___
0.6. 0.6

-2.66666667=x
nata0808 [166]3 years ago
4 0

Answer:

x = 2.67

Move variables to the right and constants to the left, Then divide and you get the answer.

Step-by-step explanation:

-1.3+4.6x=0.3+4x (given)

-1.3x + 0.6x =0.3 (subtraction property of equality)

0.6x = 1.6 (addition property of equality)

x = 2.66666666667 (division property of equality)

2.66666666667 = 2.67 (rounding to the nearest tenth)

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The sum of a number times 7 and 15 is at least -19
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Rewrite the following quadratic fiction in vertex form. Then, determine if it has a maximum or minimum and say what that value i
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5 0
4 years ago
Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0
Snowcat [4.5K]

Answer:

y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = e^{-cs}

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)}  − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 5s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 3s + 2s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[(s + 2)(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒Y(s) = \frac{3e^{-2s} }{(s + 2)(s + 3)} -  \frac{4e^{-5s} }{(s + 2)(s + 3)}

Now,

Let

\frac{1}{(s+2)(s+3)} = \frac{A}{s+2}  + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

\frac{1}{(s+2)(s+3)} = \frac{1}{s+2}  + \frac{-1}{s+3}

So,

Y(s) = 3e^{-2s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}]

⇒Y(s) = 3e^{-2s} \frac{1}{(s + 2)} -    3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ e^{-2(t-2)}  - e^{-5(t - 2)} ] - 4u₅(t) [ e^{-2(t-5)}  - e^{-5(t - 5)} ]

⇒y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

It is the required solution.

3 0
3 years ago
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