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2 years ago

It takes Wayne 1/10 hours to walk a 1/6 mile park loop. What is Wayne's unit rate, in miles per hour

1 answer:

5 0

To solve this problem, we should set up a proportion, letting x represent our unknown number of miles that Wayne walks in one hour.

1/6 mile / 1/10 hours = x miles / 1 hour

Now we use cross-products or the multiplication of the numerator of one fraction times the denominator of the other fraction, setting these two numbers equal. The resulting equation is:

1/6 = 1/10x

Finally, we must divide both sides by 1/10 to cancel it out on the right side of the equation and get the variable x alone.

x = 5/3 or 1 2/3

Therefore, Wayne walks 1 2/3 miles per hour.

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A hackberry tree has roots that reach a depth of

Degger [83]

Answer:

24.6967 meters

Step-by-step explanation:

The roots of the tree go 6 and 5 over 12 meters below the ground level.

Now, 6 and 5 over 12 meters is equivalent to 6.4167 meters.

Again the top of the tree is 18.28 meters high from the ground level.

Therefore, the total height of the tree from the bottom of the root to the top is

(6.4167 +18.28) = 24.6967 meters (Answer)

5 0

2 years ago

489 to the nearest ten

sammy [17]

8 is in the tenth position.

In order to round up, the number behind it (in this case. 9) must be one of the numbers of 5-9. Because 9 meets this requirement, you can round 8 to 9 and this will make 9 to 0.

Your answer is 490

4 0

2 years ago

Read 2 more answers

4. Suppose that Peculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply thr

Viktor [21]

Answer:

Peculiar purples would be more abundant

Step-by-step explanation:

Given that eculiar Purples and Outrageous Oranges are two different and unusual types of bacteria. Both types multiply through a mechanism in which each single bacterial cell splits into four. Time taken for one split is 12 m for I one and 10 minutes for 2nd

The function representing would be

i) $P=P_0 (4)^{t/12}$ for I bacteria where t is no of minutes from start.

ii) $P=P_0 (4)^{t/10}$ for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

a) Here P0 =3, time t = 60 minutes.

i) I bacteria P = $3(4)^{5} =3072$

ii) II bacteria P = $3(4)^{4} =768$

b) Since II is multiplying more we find that I type will be more abundant.

The difference in two hours would be

$3(4)^{10}- 3(4)^{8} =2949120$

c) i) $P=P_0 (4)^{t/12}$ for I bacteria where t is no of minutes from start.

ii) $P=P_0 (4)^{t/10}$ for II bacteria where t is no of minutes from start. P0 is the initial count of bacteria.

d) At time 36 minutes we have t = 36

Peculiar purples would be

$i) P=3 (4)^{36/12}=192$

The rate may not be constant for a longer time. Hence this may not be accurate.

e) when splits into 2, we get

$P=P_o (2^t)$ where P0 is initial and t = interval of time

7 0

2 years ago

Frank left his house at 7 a.m. and drove to the airport at a speed of 50 mph. Lance left his house at 6 a.m. and drove to the sa

Tamiku [17]

Answer:

Lance will arrive at the airport at 10:00 AM.

Step-by-step explanation:

Lance left his house at 6:00 AM, and drove to the airport, which is 220 miles away, at 55mph; therefore, the time $t$ it takes Lance to cover the distance of 220 miles is