The maximum walking speed of the Giraffe is 1.41 times greater than the maximum walking speed of the Hippopotamus
<h3>Calculating Maximum speed</h3>
From the question, we are to determine how much greater the maximum walking speed of Giraffe is to that of Hippopotamus
From the give information,
The maximum walking speed, S, is given by
S = √gL
Where g = 32ft/sec
and L is the length of the animal's leg
Thus,
For a Giraffe with a leg length of 6 feet
S = √32×6
S = √192
S = 13.856 ft/sec
For a Hippopotamus with a leg length of 3 feet
S = √32×3
S = √96
S = 9.798 ft/sec
Now, we will determine how many times greater 13.856 is than 9.798
13.856/9.798 = 1.41
Hence, the maximum walking speed of the Giraffe is 1.41 times greater than the maximum walking speed of the Hippopotamus
Learn more on Calculating Speed here: brainly.com/question/15784810
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Answer:
Step-by-step explanation:
<u>Given:</u>
- G = (11,-5)
- M = (4,-4)
- R = ?
<u>Solution</u>
<u>As per midpoint formula;</u>
- 4 = (11 + x)/2 ⇒ x + 11 = 8 ⇒ x = 3
- -4 = (-5 + y)/2 ⇒ y - 5 = -8 ⇒ y = -3
- R = (3, -3)
Do you mean
![3 \sqrt[3]{125}](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B125%7D%20)
or
![\sqrt[3]{125}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%20)
remember that
![\sqrt[n]{x^m}=x^\frac{m}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%3Dx%5E%5Cfrac%7Bm%7D%7Bn%7D)
resolve the
![\sqrt[3]{125}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D)
part first
![\sqrt[3]{125}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%3D)
![\sqrt[3]{5^3}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5%5E3%7D%3D)
so
![3 \sqrt[3]{125}=3*5=15](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B125%7D%3D3%2A5%3D15%20)
or
![\sqrt[3]{125} =5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125%7D%20%3D5)
not sure which one you mean
It is C, because the x values between the first dot and the second is 4. So you multiply the slope 4 times on the y value. The y value changes from -2 to -5, so -2 + 4*-3/4 should be equal to -5
