Approximate the value of 14−−√ to the nearest tenths place. Plot the approximation on the number line.
2 answers:
The approximation for √14 to the nearest tenth is 3.7
There are different methods to approximate a square root: the easiest is obviously the calculator, the second option is a tabular interpolation, otherwise, there is the algorithm.
Let's suppose you don't have anything but a piece of paper and a pen/pencil and you need to apply the algorithm. The steps to follow are the following:
A) Write the number you need to calculate, √14, and divide its digits into pairs starting from the units: in our case 14 is a pair.
B) Start from the first pair from the left (14) and find the greatest integer whose square is less than or equal to your pair:
3² < 14 < 4²
Write the number you found (3) in the suitable space.
C) Square the number you found in step B, write it below the pair you considered and perform a difference: 14 - 9 = 5
D) lower the second pair of the original number and write it next to the remainder found in step C. Since our number was only two-digit long, we lower 00 as if it were 14.00, and we get 500.
E) now there is the most delicate step: double the number you wrote in step B and write it below the answer: 3 · 2 = 6
Making various attempts, you need to find the greatest digit X so that the result of the multiplication 6X · X is less than or equal to the partial remainder we found in step D.
Since 14 is closer to 16 than to 9, let's start with X = 5:
65 · 5 = 325
66 · 6 = 396
67 · 7 = 469
68 · 8 = 544 > 500!!
F) Write the digit you found in the solution space, write the result of the multiplication under the partial remainder and perform the difference: 500 - 69 = 31
G) repeat step D, E, and F until you finish the pairs or until the desired approximation. In your case, you can stop here: the approximation wanted is 3.7
In the picture attached is shown the complete work.
There are different methods to approximate a square root: the easiest is obviously the calculator, the second option is a tabular interpolation, otherwise, there is the algorithm.
Let's suppose you don't have anything but a piece of paper and a pen/pencil and you need to apply the algorithm. The steps to follow are the following:
A) Write the number you need to calculate, √14, and divide its digits into pairs starting from the units: in our case 14 is a pair.
B) Start from the first pair from the left (14) and find the greatest integer whose square is less than or equal to your pair:
3² < 14 < 4²
Write the number you found (3) in the suitable space.
C) Square the number you found in step B, write it below the pair you considered and perform a difference: 14 - 9 = 5
D) lower the second pair of the original number and write it next to the remainder found in step C. Since our number was only two-digit long, we lower 00 as if it were 14.00, and we get 500.
E) now there is the most delicate step: double the number you wrote in step B and write it below the answer: 3 · 2 = 6
Making various attempts, you need to find the greatest digit X so that the result of the multiplication 6X · X is less than or equal to the partial remainder we found in step D.
Since 14 is closer to 16 than to 9, let's start with X = 5:
65 · 5 = 325
66 · 6 = 396
67 · 7 = 469
68 · 8 = 544 > 500!!
F) Write the digit you found in the solution space, write the result of the multiplication under the partial remainder and perform the difference: 500 - 69 = 31
G) repeat step D, E, and F until you finish the pairs or until the desired approximation. In your case, you can stop here: the approximation wanted is 3.7
In the picture attached is shown the complete work.
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Answer:
Step-by-step explanation:
First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is
and we can use that in a d = rt table:
d = r * t
Nina .145
Jo
Now we can fill in the distance which is 6 for both, since that is the distance where they met:
d = r * t
Nina 6 = .145
Jo 6 =
Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:
d = r * t
Nina 6 = .145 * t + 3
Jo 6 = r * t
As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then
6 = .145(t + 3) and
6 = .145t + .435 and
5.55 = .145t so
t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then
6 = r(38.379) and
r = .16 km/min
Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:
6 should equal .145454545( 38.37931034 + 3)
6 ?=? .145454545(41.37931034)
6 ?=? 6.0 so
Jo's rate is .16 km/min rounded