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Papessa [141]
4 years ago
14

The fraction 12/20 express in its simplest form is:

Mathematics
2 answers:
goldfiish [28.3K]4 years ago
8 0
Divided both of the numbers by 4, and you get the fraction of 3/5
mash [69]4 years ago
8 0
The answer is 3/5. 4 can go into both 12 and 20.
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What are the domain and range of the function
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The domain of a function is the X values 
The range of a function is the Y values  
8 0
4 years ago
Characteristics of LCM
exis [7]
The LCM (least common multiple) is the smallest positive whole number exactly divisible by two or more given whole numbers.
5 0
4 years ago
Are these two ratios proportional <br> 3/5 &amp;10/6
fomenos
First off, let's find the LCM of both 5 and 6 to easily compare.

LCM = 30

3/5 = 18/30
10/6 = 50/30

18\neq30

These ratios aren't proportional. Hope this helps!
4 0
3 years ago
Read 2 more answers
Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y e
sergiy2304 [10]

Answer:

y(t) =  - 5 {e}^{2\pi - 2t} \cos(t)

Step-by-step explanation:

The given initial value problem is

y''+4y'+5=0

y( \frac{ \pi}{2} ) = 0

y'( \frac{ \pi}{2} ) = 5

The corresponding characteristic equation is

{m}^{2}  + 4m + 5 = 0

m =  - 2 \pm \: i

The general solution becomes:

y(t) = A {e}^{ - 2t}  \cos(t)  +  B {e}^{ - 2t}  \sin(t)

We differentiate to get:

y'(t) =  - A {e}^{ - 2t}  \sin(t)  - 2 A {e}^{ - 2t}  \cos(t)  + B {e}^{ - 2t} \cos(t)   - 2B {e}^{ - 2t}  \sin(t)

We apply the initial conditions to get;

y( \frac{\pi}{2} ) = A {e}^{ - 2\pi}  \cos( \frac{\pi}{2} )  +  B {e}^{ - 2\pi}  \sin( \frac{\pi}{2} )

A {e}^{ - 2\pi} (0 )  +  B {e}^{ - 2\pi}  ( 1)  = 0

B  = 0

Also;

y'( \frac{\pi}{2} ) =  - A {e}^{ - 2\pi}  \sin( \frac{\pi}{2} )  - 2 A {e}^{ - 2\pi}  \cos( \frac{\pi}{2} )  + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} )   - 2B {e}^{ - 2\pi}  \sin( \frac{\pi}{2} )

- A {e}^{ - 2\pi} ( 1 )  - 2 A {e}^{ - 2\pi} ( 0 )  + B {e}^{ - 2\pi}( 0)   - 2B {e}^{ - 2\pi} ( 1 )  = 5

- A {e}^{ - 2\pi}     - 2B {e}^{ - 2\pi} = 5

But B=0

- A {e}^{ - 2\pi}  = 5

A = 5{e}^{2\pi}

Therefore the particular solution is

y(t) =  - 5 {e}^{2\pi} {e}^{ - 2t}  \cos(t)  +  0 \times {e}^{ - 2t}  \sin(t)

y(t) =  - 5 {e}^{2\pi - 2t} \cos(t)

4 0
4 years ago
3 | 2x – 3| -5 = 4<br><br> Absolute value equations
anyanavicka [17]

Answer:

3

Step-by-step explanation:

  1. 3 |2x -3| -5=4
  2. 3 |2x -3| =4+5
  3. 2x-3=9÷3
  4. 2x=3+3
  5. x=6÷2
  6. x=3
8 0
4 years ago
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