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Dovator [93]
3 years ago
8

Solve 3/5x=15 Please

Mathematics
1 answer:
Ilya [14]3 years ago
5 0

Answer:

dude

if 3 = 15

5x = 5.5=25

Step-by-step explanation:

hope it helps

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Sorry forgot to post the problem here it is
vovikov84 [41]

Answer:

55

Step-by-step explanation:

m angle A and m angle C are equal sine two of those sides are equal. You would do 3x + 40 = x +50

You subtract x, leaving it to be 2x + 40 = 50

Subtract the 40, leaving it be 2x = 10

Divide both sides by 2, which leaves x = 5

and then you would do 3(5) + 40, which is 55

4 0
3 years ago
HELP!!!!<br> 2x+y= 20<br> 3x+4y= 40
bagirrra123 [75]

Answer:

(8,4)?

Step-by-step explanation:

3 0
3 years ago
Jina received a $70 gift card for a coffee store. She used it in buying some coffee that cost $7.51 per pound. After buying the
dem82 [27]

Answer: 3 pounds

Step-by-step explanation:

The equation can be set up according to slope form: y=mx+b where x represents cost per round.

$70 = $7.51x + $47.47

Subtract 47.47 to both sides: $22.53 = $7.51x

Divide both sides by 7.51: 3 = x

4 0
3 years ago
Read 2 more answers
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The answer is c Dhsbnsndgdh
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The The Laplace Transform of a function , which is defined for all , is denoted by and is defined by the improper integral , as
guapka [62]

Answer:

a. L{t} = 1/s² b. L{1} = 1/s

Step-by-step explanation:

Here is the complete question

The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0

Solution

a. L{t}

L{t} = ∫₀⁰⁰e^{-st}t

Integrating by parts  ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = e^{-st} and v = \frac{e^{-st}}{-s} and du/dt = dt/dt = 1

So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w

So,  ∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

∫₀⁰⁰e^{-st}t =  [\frac{te^{-st}}{-s}]₀⁰⁰ -  ∫₀⁰⁰ \frac{e^{-st}}{-s}

= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + \frac{1}{s} [\frac{e^{-st} }{-s}]₀⁰⁰

= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]

= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]

= -1/s[(0 - 0] - 1/s²[0 - 1]

= -1/s[(0] - 1/s²[- 1]

= 0 + 1/s²

= 1/s²

L{t} = 1/s²

b. L{1}

L{1} = ∫₀⁰⁰e^{-st}1

= [\frac{e^{-st} }{-s}]₀⁰⁰

= -1/s[exp(-∞s) - exp(-0s)]

= -1/s[exp(-∞) - exp(-0)]

= -1/s[0 - 1]

= -1/s(-1)

= 1/s

L{1} = 1/s

6 0
3 years ago
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