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gladu [14]
2 years ago
15

Using the distributive property what is 4(2x -1) in simplest form?

Mathematics
2 answers:
Marina86 [1]2 years ago
4 0

Answer:

8x -4

Step-by-step explanation:

ollegr [7]2 years ago
4 0
4(2x-1)
8x-4
This is as much as you can simplify it because the 8x cannot go into 4 but if it was 4x then it could go into the 8
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Please help. Will mark brainliest if correct. ( links will be banned)
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Answer:

answer 2

Step-by-step explanation:

I have no clue guess you'll just have to trust me

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Coordinate and it's reflection over the x- axis be an even number
PilotLPTM [1.2K]
Depends on the original point
4 0
3 years ago
A tennis player keeps track of the number of successful first serves he makes. During the first 8 service points of a game, only
Andru [333]

Answer:

He needs 7 more consecutive successful first serves to raise his first serve percentage to 60%.

Step-by-step explanation:

After n consecutive serves, his total number of serves is going to be n+8, since he has already served 8 times. In the best case, his number of successful first serves is n+2.

His percentage of succesful first serves is the division of the number of succesful first serves divided by the total number of serves. So

P = \frac{n+2}{n+8}

We want P = 0.60. So

0.6 = \frac{n+2}{n+8}

n+2 = 0.6*(n+8)

n + 2 = 0.6n + 4.8

n - 0.6n = 4.8 - 2

0.4n = 2.8

n = \frac{2.8}{0.4}

n = 7

He needs 7 more consecutive successful first serves to raise his first serve percentage to 60%.

5 0
3 years ago
Can someone help me ?
lozanna [386]

Answer:

10 > v (if you don't trust me, check it yourself. Subsitute v for 10 in the first equation.

Step-by-step explanation:

1) 7 > v - 3

2) add 3 to both sides

3) so 7 plus 3 equals 10

So you basically do this

7 > v - 3 (ad 3 to both sides)

You get 10 > v (which is as simplified as possible)

Hope this helps! :)

8 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
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