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2 years ago

Find angle c° and d°.

Mathematics

1 answer:

Vladimir79 [104]2 years ago

7 0

Answer:

Step-by-step explanation:

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Which function is non linear y-4x=1 y=2+6x^4 x-2y=7

ioda

Answer:

Step-by-step explanation:

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2 years ago

What are the solutions of x^2-4x+5=0

Gnoma [55]

Answer:

Factor {x}^{2}-4x-5x

−4x−5.

(x-5)(x+1)=0(x−5)(x+1)=0

2 Solve for x.

x=5,-1x=5,−1

Step-by-step explanation:

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2 years ago

Read 2 more answers

What is the area of the trapezoid

Zarrin [17]

I'm not sure because u don't have a pic of the trapezoid.

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2 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

What is additive inverse of the check?

elena-14-01-66 [18.8K]

Answer:

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7 0

3 years ago