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3 years ago

Find the unknown side of the right triangle below. Round to the nearest tenth

Mathematics

2 answers:

dmitriy555 [2]3 years ago

4 0

Answer:

7.1

Step-by-step explanation:

5^2 + 5^2 = c^2

25 + 25 = c^2

50 = c^2

7.1 = c

shtirl [24]3 years ago

3 0

Answer:

B) 7.1 ft

Step-by-step explanation:

Two ways to go about this:

<u>Method 1:</u>

Recognize that since the side lengths are equal to each other, then the hypotenuse must be equal to 5√2 or about 7.1 feet since the triangle is a 45-45-90 triangle.

<u>Method 2:</u>

By using the Pythagorean Theorem, a^2+b^2=c^2, we have 5^2+5^2=c^2 --> 25+25=c^2 --> 50=c^2 --> 7.1≈c

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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 $0.127 \frac{cm}{min}$

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