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hammer [34]
3 years ago
7

Cho f(t)=(t-4)u(t-2)phep bien doi laplace la F(s)=e^-2s/s^2-2e^-2s/s62

Mathematics
1 answer:
FinnZ [79.3K]3 years ago
4 0

Step-by-step explanation:

cho f(t)=(t-4)u(t-2)phep bien doi laplace la

F(s)=e^-2s/s^2-2e^-2s/s62

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What is the equation written in vertex from of a parabola with a vertex of (4, –2) that passes through (2, –14)?
vichka [17]

Answer:

y = -3(x - 4)² - 2

Step-by-step explanation:

Given the vertex, (4, -2), and the point (2, -14):

We can use the vertex form of the quadratic equation:

y = a(x - h)² + k

Where:

(h, k) = vertex

a  =  determines whether the graph opens up or down, and it also makes the parent function <u>wider</u> or <u>narrower</u>.

  • <u>positive</u> value of a = opens <u><em>upward</em></u>
  • <u>negative</u> value of a = opens <u><em>downward</em></u>
  • a is between 0 and 1, (0 < a < 1) the graph is <u><em>wider</em></u> than the parent function.
  • a > 1, the graph is <u><em>narrower</em></u> than the parent function.

<em>h </em>=<em> </em>determines how far left or right the parent function is translated.

  • h = positive, the function is translated <em>h</em> units to the right.
  • h = negative, the function is translated |<em>h</em>| units to the left.

<em>k</em> determines how far up or down the parent function is translated.

  • k = positive: translate <em>k</em> units <u><em>up</em></u>.
  • k = negative, translate <em>k</em> units <u><em>down</em></u>.

Now that I've set up the definitions for each variable of the vertex form, we can determine the quadratic equation using the given vertex and the point:

vertex (h, k): (4, -2)

point (x, y): (2, -14)

Substitute these values into the vertex form to solve for a:

y = a(x - h)² + k

-14 = a(2 - 4)²  -2

-14 = a (-2)² -2

-14 = a4 + -2

Add to to both sides:

-14 + 2 = a4 + -2 + 2

-12 = 4a

Divide both sides by 4 to solve for a:

-12/4 = 4a/4

-3 = a

Therefore, the quadratic equation inI vertex form is:

y = -3(x - 4)² - 2

The parabola is downward-facing, and is vertically compressed by a factor of -3. The graph is also horizontally translated 4 units to the right, and vertically translated 2 units down.

Attached is a screenshot of the graph where it shows the vertex and the given point, using the vertex form that I came up with.

Please mark my answers as the Brainliest, if you find this helpful :)

8 0
2 years ago
Name two solutions to y = 3x - 5
wlad13 [49]

Answer:

(1,-2) and (0,-5) (There are an infinite amount more)

Step-by-step explanation:

The easiest way to find solutions is to plug in an x value. Let's try 1:

y = 3(1) - 5⇒y = 3 - 5⇒y = -2

(1,-2)

Let's try 0:

y = 3(0) - 5⇒y = 0 - 5 ⇒y = -5

(0,-5)

6 0
3 years ago
A store sells candles.
maria [59]

Answer:

25%

Step-by-step explanation:

If you start at 45% and it reduces by 20% you are just doing 45-20 which equals 25. (Please let me know if im wrong)

7 0
3 years ago
Consider the function f(x) = √5x + 15 for the domain [-3, ∞).Find f-¹(x), where f-1 is the inverse of f.Also state the domain of
Ymorist [56]

Answer: We have to find the inverse function of the following:

f(x)=\sqrt{5x+15}\rightarrow[-3,\infty)\Rightarrow(1)

The Inverse of the (1) is as follows:

\begin{gathered} f(x)=y=\sqrt{5x+15} \\  \\ \text{ Switch: }x\text{  and }y\text{ and solve for }y(x)\text{ } \\  \\  \\ y=\sqrt{5x+15}\rightarrow x=\sqrt{5y+15} \\  \\  \\ x^2=5y+15 \\  \\  \\ x^2-15=5y \\  \\  \\ y=\frac{x^2}{5}-\frac{15}{5} \\  \\  \\  \\ y=f^{-1}(x)=\frac{x^2}{5}-3 \\  \\  \\ f^{-1}(x)=\frac{x^{2}}{5}-3 \end{gathered}

The domain of the inverse function is:

x\in(-\infty,+\infty)

3 0
9 months ago
An AC generator produces an alternating current with a peak value of 18 V. The resistance of the lamp is 4 ohms. What is the eff
leva [86]

The effective current when an AC peak voltage of 18V is applied to a lamp with 4 ohm resistance is 3.18 A.

<h3>What is current?</h3>

Current is the flow of electrons through a conductor when voltage is applied across the conductor.

Given the peak voltage of 18 V, hence:

V_{rms}=\frac{Peak\ value}{\sqrt{2} } =\frac{18}{\sqrt{2} } =12.73 V\\\\I_{rms}=\frac{V_{rms}}{resistance } =\frac{12.73}{4 } =3.18 A\\

The effective current when an AC peak voltage of 18V is applied to a lamp with 4 ohm resistance is 3.18 A.

Find out more on current at: brainly.com/question/24858512

7 0
2 years ago
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