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postnew [5]
1 year ago
10

S =The maximum walking speed S, in feet per second, of an animal can be modeled by the equationVgL, where g = 32 ft/sec and L is

the length, in feet, of the animal's leg. To the nearesthundredth, how many times greater is the maximum walking speed of a giraffe with a leg length of 6feet than a hippopotamus with a leg length of 3 feet?1.41 times2.00 times0.71 times4.06 times
Mathematics
1 answer:
grin007 [14]1 year ago
3 0

1.41 times

1) Well, let's get started by writing out the equation:

S=\sqrt[]{gL}

2) Let's find out the maximum walking speed of a giraffe, plugging into that the given data: Giraffe Leg Length: 6

\begin{gathered} S_{\text{giraffe}}=\sqrt[]{32\times6} \\ S_{\text{giraffe}}=\sqrt[]{192}\approx13.86 \end{gathered}

And the hippotamus:

S_{\text{hippo}}=\sqrt[]{32\times3}\Rightarrow\sqrt[]{96}\approx9.80

And now how many times greater by dividing the giraffe's speed by the hippopotamus speed:

\frac{13.86}{9.8}\approx1.41

Hence, the answer is 1.41 times.

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What is the volume of a sphere with a radius of 4
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Step-by-step explanation: To find the volume of a sphere, start with the formula for the volume of a sphere which is shown below.

V = \frac{4}{3}\pi  r^{3}

Here, we are given that our sphere has a radius of 4 units.

So plugging into the formula, we have (\frac{4}{3})(\pi)(4 units)^{3}.

Start by simplifying the exponent.

(4 units)³ is equal to (4 units) (4 units) (4 units) or 64 units³.

So we have (\frac{4}{3})(\pi)(64 units^3}).

Next, we multiply (4/3)(64) which can

be thought of as (4/3)(64/1)

So multiplying across the numerators and across the denominators,

we have \frac{256}{3} \pi units^{3}.

5 0
2 years ago
A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per
Tems11 [23]

Answer:

The pressure is changing at \frac{dP}{dt}=3.68

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} and we want to find at what rate is the pressure changing.

The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

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3 years ago
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Answer:

x = -1

Step-by-step explanation:

f(x) = g(x) when x = -1

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Answer:

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3 years ago
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