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melisa1 [442]
3 years ago
7

Which number is equivalent to 17.02? 17.020 , 17.22 , 17 , 17.002

Mathematics
1 answer:
Irina18 [472]3 years ago
3 0

Answer:

Step-by-step explanation:

17.020

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Point A is located at (1,5), and point M is located at (-1,6). If point M is the midpoint of AB find the location of point B.
Lina20 [59]

Answer:

(0,5.5)

Step-by-step explanation:

see attached image :)

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F(x)=5x+7 and g(x)=2x+3 find a) `(f+g)(x)` b) `(f-g)(x)`
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X + 1 by x = 99 find the value of 100x by 2x square + 102x + 2

1

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Please help! ill give you a good review if you give me the correct answer!
earnstyle [38]

Answer: H

Step-by-step explanation:

The value would equal to -3 1/2 it also would equal to a 3 1/2 with a different answer. But if done completely different way of completion it would also equal to 1evaluate 1 Factor so it would equal to 1.

But in your case its just H  

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3 years ago
I need help hurry plzzz!!
JulijaS [17]

Answer:

Harry swims at 1.6 laps per minute and Larry swims at 1.8 laps per minute. Henry swims at a slower rate than Larry.

Step-by-step explanation:

i dont know for sure if this is right or not but i think it is. i mean for Henry the only number i have is 1.6 and so if that is going to be multiplied by x which is his minutes i would guess that that is the rate he is swimming at per minute. and as for larry, i took 4.5 (which is his laps) and divided it by 2.5 (which is his time) to get how many laps per minute he swims. i got 1.8. 1.6 is slower than 1.8 so therefore, i concluded that harry swam at a slower rate than larry.

7 0
3 years ago
The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope
lesya692 [45]

(1) D

L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt

Integrate by parts, taking

u = t \implies \mathrm du=\mathrm dt

\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}

Then

L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}

L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}

(2) A

L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt

L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}

L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}

7 0
3 years ago
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