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Cerrena [4.2K]
2 years ago
6

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
Mathematics
1 answer:
Likurg_2 [28]2 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)

6 0
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.

5 0
2 years ago
The Laplace Transform of a function f(t), which is defined for all t &gt; 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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