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ANEK [815]
3 years ago
10

Which is a factor of 24x2 − 44x + 12? 3 3x + 3 3x − 2 2x − 3

Mathematics
2 answers:
Kipish [7]3 years ago
7 0
24x²-44x+12 = 4(2x-3)(3x-1)

<span>2x − 3 </span>
Ludmilka [50]3 years ago
4 0

Answer:

D) 2x − 3.

Step-by-step explanation:

Given:  24x² − 44x + 12.

To find : Which is a factor .

Solution : We have given

24x² − 44x + 12.

On taking 4 common to all

4 (  6x² − 11x + 3).

Now Factor the term

6x² − 11x + 3.

6x² −9x - 2x + 3

On taking common 3x from first two terms and -1 from last two terms .

3x ( 2x -3) -1 ( 2x - 3).

On grouping

(3x -1) ( 2x -3)

4 (3x -1) ( 2x -3) all are factors .

Therefore, D) 2x − 3.

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Mr and mrs. Dorsey and their three children are fling to Springfield. The cost of each ticket is $179. Estimate how much the tic
Soloha48 [4]
<span>First figure out how many people there are all together. So mrs and mr is 2 there three children so 2+3=5 so next round 179 to 200 and multiply 200x5. To do that you can use the mental math way witch would be to take the whole number out of 200 and make it 2. Then take the 5 in 200x5 and multiply 2x5. Then add the 2 zeros there were. And your estamated answer would be 1,000.
2+3=5
179=200
200x5
2x5=10
1,000
To figure out your real answer you will have to take 179 and multiply it by 5 so it will look like this:
 34
179
x
   5
895
Soto conculde the answer would be:

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Hope this helps!</span>
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3 years ago
What is the prime factorizatiom of 110
Korvikt [17]
2 x 5 x 11 Is it I think. What do you think?
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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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Step-by-step explanation:

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