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KatRina [158]
3 years ago
15

If f = {(2,5), (3, 2) (4, 6), (5, 1), (2, 2)), then f is a function.

Mathematics
1 answer:
KiRa [710]3 years ago
8 0

Answer:

false

Step-by-step explanation:

f is not a function because you can see that two of the points are (2,5) and (2,2). A function can only have one corresponding y-coordinate for every x-coordinate and since this is not the case here, f is not a function.

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Answer:

Step-by-step explanation:

First equation

x + y ≥ 10

Second equation

5.50x + 3y ≤ 45

a possible soution can be

x = 4 and y= 6

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Please help!!! This is one question with a couple parts I am extremely confused and it would be amazing if you could show steps!
Dovator [93]

Answer:

  a)  |n -11| = 5

  b)  n ∈ {6, 16}

Step-by-step explanation:

The wording of the question is ridiculous. We assume it is intended to read, "The distance between two numbers is 5. One of the numbers is 11. What are the possibilities for the other?"

a) The distance between a number (n) and 11 can be written as ...

  |n -11|

Since we want that distance to be 5, we can write the equation ...

  |n -11| = 5

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b) The equation resolves to two:

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Adding 11 to both sides of both equations gives ...

  • n = 16
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The two solutions are n=6 and n=16.

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<em>Comment on the question statement</em>

Increasingly, we see curriculum materials written in Pidgin English or where the words have a meaning different from that understood by a native English speaker. It appears you are the lucky recipient of such materials, so must do occasional "interpretation". Here, it seems that "two time a number" is intended to mean "two numbers."

4 0
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In a math problem what is 12+6=3+5
givi [52]

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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
(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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3 years ago
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