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3 years ago

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The snow on a market roof has a density of 100kg/m3 and is 0.2 deep. That roof is 20 m wide and 40 m long. What is the load of t

he snow

1 answer:

Likurg_2 [28]3 years ago

7 0

16,000 Kg of snow. 20 x 40 x 0.2 = 160 x 100 = 16,000

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Choose the point-stope form of the equation below that represents the line that passes through the point (-1, 6) and has a slope

Keith_Richards [23]

Answer:

y-6 = -3(x+1)

Step-by-step explanation:

The point-slope form of a line is the following:

y-yo = m(x-xo), where 'm' is the slope and (xo, yo) is any point where the line passes through.

In this case, m=-3 and (xo, yo) = (-1, 6).

Therefore: y-yo = m(x-xo) = y-6 = -3(x+1)

In conclusion, the point-slope form of the equation that represents the line that passes through the point (-1, 6) and has a slope of -3 is:

y-6 = -3(x+1)

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3 years ago

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IrinaVladis [17]

Answer:

a-2/3b

Step-by-step explanation:

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Kara has 21 collectible buttons to sell. She sells the small buttons for $3 and the large buttons for $4, and earns a total of $

Serga [27]

We know that there are 21 buttons total. x is the number of small buttons and y is the number of large buttons.

Therefore, our first equation should be x+y = 21.

We know the small buttons sell for 3 dollars, meaning 3x represents how much is made from the small buttons. The larger buttons sell for 4 dollars, meaning 4y is how much is made from the large buttons.

We also know that the total of the buttons sold is 68 dollars. Therefore 3x + 4y = 68 is our second equation.

From this information, the answer is the first set.

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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

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

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

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

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

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3 years ago

Need help this is the rest of my points.

pickupchik [31]

Answer: C

Step-by-step explanation:

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3 years ago