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masya89 [10]
2 years ago
15

Could anyone answer this math question for me?

Mathematics
1 answer:
Rudik [331]2 years ago
6 0

Answer:

11 ft tall

Step-by-step explanation:

since the stop sign shadow just added 1 foot from the original height then do the opposite when finding the height of the school bus

12 ft - 1 foot = 11 ft

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Cleatus pulled into the neighborhood gas station to fill his car with gas. The 91-octane gas is selling for $2.08 per gallon. He
Dafna1 [17]

Answer: $47.60

Step-by-step explanation: First, we will multiply $2.08 by 20 because we want to know how much it will cost for 20 gallons of gas.

$2.08 x 20 = $41.60

Now, just add the $6.00 car wash to the total amount of gas which is $41.60.

$41.60 + $6.00 = $47.60

So, the total amount of his purchase was $47.60

7 0
3 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
What is the value of x in the equation 5(3x + 4) = 23
abruzzese [7]

5(3x+4)=23

15x+20=23

15x=23-20

15x=3

x=3/15

x=1/5=0.2

4 0
2 years ago
Please Help I Will Mark Brainly!!!
Lesechka [4]

Answer:


d because the fomula is always (x,y)


4 0
3 years ago
Write an equation for the graph below in terms of x
monitta

Answer:

x = y/2 - 1/2

Step-by-step explanation:

you can use the slope-intercept form of a line first, then change to terms of 'x'

y = mx + b    where 'm' is the slope and 'b' is the y-intercept

if we use two points on the line, such as (-1, -1) (0, 1), we can find the slope

m = 1-(-1) / 0-(-1)

m = 2/1

b = 1

Slope-Intercept form:  y = 2x + 1

in terms of 'x':

2x = y - 1

x = y/2 - 1/2

6 0
2 years ago
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