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jasenka [17]
3 years ago
15

Four team members are running a 3 mile team relay. How many miles does each member run?

Mathematics
1 answer:
SSSSS [86.1K]3 years ago
6 0
3 miles / 4 people =  3/4 mile each
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a rock band has 5 members and, 2/5 of the members play string instruments. Also, 0.4 of the members sing.Does the band have the
Bad White [126]
Convert 2/5 to a decimal and you get .4, or 40%. 40% of the band members also sing, so yes, the band has the same number of string instrument players and singers.
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Whats greater 9.354 or 9.012
babymother [125]

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9.354

Step-by-step explanation:

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Write the equation of the circle graphed below.
Darina [25.2K]

Answer:

Your equation is (x - 1)^2 + (y +1)^2 = (\frac{1}{2}  )^2

Step-by-step explanation:

Well, the center origin of the circle is given (h,k) =  (1,-1).

We have to find our radius as they gave us a point. from origin to the edge of the circle.

Using the formula: (x - h)^2 + (y - k)^2 = r^2

Plug in our (h,k) = (1,-1) and (x,y) =  (0.5,-1) to solve for radius.

(x - h)^2 + (y - k)^2 = r^2

(0.5 - (1))^2 + (-1 - (-1))^2 = r^2

(-0.5)^2 + (0)^2 = r^2

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2 years ago
Which net represents this triangular prism?
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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}}

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3 years ago
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