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shutvik [7]
3 years ago
8

May I know this answer

Mathematics
1 answer:
Anettt [7]3 years ago
6 0
I think B should be the top one
You might be interested in
as a sales person at Trending Card Unlimited, Justin receives a monthly base pay plus commission on all that he sells. If he sel
Amiraneli [1.4K]

Answer:

  $1025

Step-by-step explanation:

We can use the 2-point form of the equation of a line to write a function that gives Justin's salary as a function of his sales.

We start with (sales, salary) = (400, 500) and (700, 575)

__

The 2-point form of the equation of a line is ...

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  salary = (575 -500)/(700 -400)(sales -400) +500

  salary = 75/300(sales -400) +500

For sales of 2500, this will be ...

  salary = (1/4)(2500 -400) +500 = (2100/4) +500 = 1025

Justin's salary after selling $2500 in merchandise is $1025.

5 0
3 years ago
Plz help im an idiot
MrMuchimi

Answer:

3.)138

4.) 149

Step-by-step explanation:

you add the two inside measures together.

5 0
2 years ago
Read 2 more answers
Finding surface area if an triangular prism
SVEN [57.7K]
1)find the area of a triangle. Multiply it by 2 or 4 depending on if they are the same size
2)find the area of the base
3)find the area of the other two sides if you haven’t already.
4) add them all together
8 0
3 years ago
The measure of arc EF is —
Dimas [21]

30 plus EF = 180

Ef = 150

circumscribed angle plus the central angle is 180

7 0
2 years ago
Read 2 more answers
(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
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