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garri49 [273]
3 years ago
9

The value of 21b - 32 + 7 b - 20 b is​

Mathematics
2 answers:
navik [9.2K]3 years ago
7 0
8b-32, 8b=32, b=4 it says I need 20 characters sooooo
Aleonysh [2.5K]3 years ago
6 0
8b - 32

Hope it’s right
Best luck with your studying
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A train on a straight track goes in the positive direction for 6.7 km , and then backs up for 4.0 km . Is the distance covered b
RSB [31]

Answer:

Distance is greater than the displacement.

Step-by-step explanation:

We are given that a train on a straight track goes in the positive direction.

We have to find that distance covered by the train is greater than, lees than or equal to its displacement.

Train covered distance in positive direction=6.7 km

Train covered distance in back=4.0 km

Distance covered by the train=6.7+4=10.7 Km

Because distance is the length of path covered by object.

Displacement =Final position -Initial position

Displacement =6.7-4=2.7 km

        10.7 > 2.7

Distance > Displacement

Hence, Distance is greater than the displacement.

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7 0
3 years ago
(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}

3 0
3 years ago
Tori needs at least $335 dollars to buy a new bat. She has already saved $125. She
Katen [24]

Answer:

she would need to work 14 hours

Step-by-step explanation:

if you decrease 125 from 335 you would need to find out how to get 210 from 15 if you multiply 14 and 15 together you would end up with 210

3 0
3 years ago
I need help please ​
snow_tiger [21]

Answer:2nd one

Step-by-step explanation:

4 0
3 years ago
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