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Archy [21]
3 years ago
15

What is the value of 3/2 + 4/2?​

Mathematics
2 answers:
earnstyle [38]3 years ago
6 0

Answer:

7/2 or 3 1/2

hope this helps

have a good day :)

Step-by-step explanation:

Finger [1]3 years ago
6 0

Answer:

7/2

Step-by-step explanation:

3/2 + 4/2 = 7/2

Hope this helps!!

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Write the word sentence as an inequality. Then solve the inequality.
ss7ja [257]

Answer:

x/7<-3

x<-21

Step-by-step explanation:

x/7<-3

multiply the seven to get rid of x denominator

x<-21

6 0
3 years ago
What is an equation of the line that passes through the point (3, -8) and has slope -2?
BabaBlast [244]

Answer:

Step-by-step explanation:

slope of -2 means the slope is -2/1

from (3, - 8) you can find the y intercept by going backwords until x = 0

(2,-6)

(1,-4)

(0,-2)

so the y intercept is -2

y = -2x - 2

6 0
2 years ago
Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

3 0
3 years ago
How to help my child study times tables
Alex787 [66]

Everyday, make her/him learn 3-4 tables. After 3 days, give him/her a 'prize' for it. If that does not work, try teaching him 1 by 1 and memorize it.

6 0
3 years ago
Can someone please help me this is geometry I don’t understand I have like more than an hour trying to figure this but I cancele
Aleks04 [339]

Answer:

5, last option

Step-by-step explanation:

Sides to be calculated are equal as both are opposite same angles, are adjacent to common hypotenuse of right triangles

  • 3x+1=2x+6
  • x=6-1
  • x=5
6 0
3 years ago
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