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torisob [31]
2 years ago
10

Simplify. y 5 ÷ y 3 • y 2

Mathematics
1 answer:
icang [17]2 years ago
5 0

Answer:

30y

Step-by-step explanation:

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What are the solutions of the quadratic equation? 4x^2+34x+60=0
insens350 [35]

Answer:

Option 3) -6,-5/2

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

4x^{2} +34x+60=0  

so

a=4\\b=34\\c=60

substitute in the formula

x=\frac{-34(+/-)\sqrt{34^{2}-4(4)(60)}} {2(4)}

x=\frac{-34(+/-)\sqrt{196}} {8}

x=\frac{-34(+/-)14} {8}

x_1=\frac{-34(+)14} {8}=-\frac{20}{8}=-\frac{5}{2}

x_2=\frac{-34(-)14} {8}=-6

6 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
3 years ago
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