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gizmo_the_mogwai [7]
3 years ago
5

Please help me idk this

Mathematics
1 answer:
madreJ [45]3 years ago
8 0

Answer:

660in

Step-by-step explanation:

Rectangle:

A=l*w

A=20*30=60

Triangle:

A=b*h/2

A=40*30/2=600

600+60=660

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17. Simplify (3 × 22) ÷ 6 + [28 – (4)2] = ?
Effectus [21]

Answer:

31

Step-by-step explanation:

(3*22=66) / 6 + [28n - (4)2=20]

66 /6 = 11+ [20]

11+20= 31

7 0
3 years ago
I need this answer asap
antoniya [11.8K]

Answer:

240

Step-by-step explanation:

5 0
3 years ago
Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat
nirvana33 [79]

Answer:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

V = \pi r^2 h

For this case we know that r=5m represent the radius, h = 10m the height and the rate given is:

\frac{dV}{dt}= \frac{100 L}{min}

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

And solving for \frac{dh}{dt} we got:

\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}

We need to convert the rate given into m^3/min and we got:

Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}

And replacing we got:

\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}

And that represent 0.127 \frac{cm}{min}

5 0
4 years ago
Which inequality best represents the situation in which you must be at least 42 inches tall to ride the roller coaster
Romashka-Z-Leto [24]
42  \leq x

6 0
3 years ago
Simplify problem if possible
sergeinik [125]
\left(\frac{(x^4y^{-2})^4}{(2x^2)^2\cdot2x^0y^3}\right)^2=\\
\left(\frac{x^{16}y^{-8}}{4x^4\cdot2y^3}\right)^2=\\
\left(\frac{x^{16}y^{-8}}{8x^4y^3}\right)^2=\\
\left(\frac{x^{12}y^{-11}}{8}\right)^2=\\
\frac{x^{24}y^{-22}}{64}
3 0
4 years ago
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