A cheetah runs 87 meters in 3 seconds. what is the cheetahs unit rate of speed?

Mathematics

1 answer:

Sunny_sXe [5.5K]3 years ago

6 0

Well if we do this:
87m/3s
we get:
29m/s

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Aleks04 [339]

Answer:

5.8

Step-by-step explanation:

Side KI is proportional to HF, and as we know GH is proportional to JK, we can make a proportion. x/31 = 8/43, 43x = 248, so x = 5.767, which rounds to 5.8

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2 years ago

Wayne took 2 hours to cover 1/3 of a trip. He covered the remaining 60 miles in 2 hours. Find his average speed for the entire t

faust18 [17]

Wayne traveled 90 miles total.
60 miles was 2/3 of the distance.
30 miles is 1/3.
90 miles divided by 4 hours.
90/4
22.5 or 22 1/2 mph

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3 years ago

A collection of 20 coins made up of only nickels, dimes, and quarters has a total value of $3.35. If the dimes were nickels, the

Kay [80]

Answer: The required number of quarters in the collection is 11.

Step-by-step explanation: Given that a collection of 20 coins made up of only nickels, dimes and quarters has a total value of $3.35.

If the dimes were nickels, the nickels were quarters and the quarters were dimes, the collection of coins would have a total value of $2.75.

We are to find the number of quarters in the collection.

Let x, y and z represents the number of nickels, dimes and quarters respectively in the collection.

We will be using the following values of nickels, dimes and quarters in form of dollar :

1 nickel = $ 0.05, 1 dime = $ 0.10 and 1 quarter = $0.25.

Then, according to the given information, we have

$x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Substituting the value of x from equation (i) in equations (ii) and (iii), we have

$(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)$

and

$y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)$

Comparing the values of y from equations (iv) and (v), we get

$47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.$