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3 years ago

Which expression is equivalent to 11 - (-3 5/8) ?

Mathematics

1 answer:

scoray [572]3 years ago

7 0

11 - (-3 5/8)

-negative sign and -negative sign= +positive sign

11+3 5/8

answer:

D. 11 + 3 5/8

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Answer:he 36

Step-by-step explanation:

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What is 154 divided by 12? Type in the letter of the correct answer: a) 12⅚ b) 12⅙ c) 12¼ d) 12¾ e) 12

shutvik [7]

Answer:

Step-by-step explanation:

154÷12=12 5\6

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3 years ago

Find the value of x: 7x + 15 = 38

borishaifa [10]

Answer:

23/7

Step-by-step explanation:

7x + 15 = 38

subtract 15 from both sides : 7x+15-15=38-15

7x=23

divide both sides by 7: 7x/7 = 23/7

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3 years ago

Consider the hypotenuse of a right triangle with two legs that are two inches and one inch long, respectively. The length of the

Olenka [21]

Answer:

Step-by-step explanation:

Pythagoras formula for the relation of the side lengths of a right-angled triangle :

c² = a² + b²

c is the Hypotenuse (the side opposite of the 90 degree angle). a and b are the "legs".

so, we have here

c² = 1² + 2² = 1 + 4 = 5

c = sqrt(5) = 2.236067977...

so, answer option c is too big (values bigger than 3).

answer options d and e are too small (values smaller than 2).

9/4 = 2.25

10/4 = 2.5

this is also too big .

that leaves us with b

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that is correct

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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

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(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

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

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

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

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3 years ago