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3 years ago

Can someone help me out? (serious answers only)

Mathematics

1 answer:

Vitek1552 [10]3 years ago

5 0

Answer:

Step-by-step explanation:

the diameter is 1.8 in.

diameter to circumference is multiplying the diameter by pi

1.8*= about 5.66 (Rounded to the nearest hundredth)

the 5.66 in. is going around 15 times

multiply 5.66 and 15

5.66*15=84.9

the best answer would be A.

I don't know if this is right if it is can u mark me brainliest

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Orlov [11]

Answer:

Reflected over the x-axis, vertically shrunk by a factor of 1/5, moved right 4 units and up 2 units.

Step-by-step explanation:

Parent Graph: f(x) = a(bx - h)² + k

a is vertical stretch (a > 1) or shrink (a < 1) and reflection over x-axis

b is horizontal stretch or shrink and reflection over y-axis

h is horizontal movement left (if positive) or right (if negative)

k is vertical movement up (if positive) or down (if negative)

Now that we have our rules, we simply apply it to y = -1/5(x - 4)² + 2

a = -1/5, so reflection over x-axis and vertical shrink of 1/5 (1/5 < 1)

Nothing has changed b, so no horizontal stretch/shrink

h = -4, so horizontal movement right 4 units

k = 2, so vertical movement up 2 units

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4 years ago

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nirvana33 [79]

Answer:

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

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

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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Answer:

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