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Sagot :
D. 100
When you have a decimal (such as 5.6), each number is a representation of a place. (Tenths, Hundreds, Thousands, etc.)
So in a random example 0.123 - 1 is the tenths, 2 is the hundreds, 3 is the thousands.
Multiplying any 2 digit (or less) decimal by 100 (or more) changes the location of that decimal by moving it to the right by the given "places" (1 for 10, 2 for hundred, 3 for thousands).
So 5.6 becomes 560.0 when multiplied by 100. We have moved the decimal two places to the right, because we are multiplying by the hundreds place.
As another example, let's say we had a decimal of 0.561. We are now in the "thousands" place in decimals, so the only way to get rid of the decimal is to multiple the number by 1000 (or more), because the decimal is long enough to hit the "thousands" place.
At the same time, with the decimal "5.6", we could multiply it by 10, because it only has one decimal and therefore is in the "tenths" place. That makes it "56" instead of "560".
So why do we have to use 100 instead of 10? Because the question requires 1 answer for the entire equation, and in the equation is the number "0.12", which hits the "hundreds" place.
Why 10 or 100 and not 11 or 134 (ie any number greater than these 10 and 100)? Two reasons - first it is the lowest number possible to reach our goal. Second is because using round numbers by 10s preserves the original decimal easily for our eyes.
If we did 5.6 x 134 (for example) we end up with 750.4. If I handed that to Bob, he'd have to do some complicated math to get the answer back to the original equation. Our math would not be incorrect (as long as we used 134 for every number in the equation), but it would be annoying! And it makes us more at risk to mess up our math!
Let me know if I can clarify anything further.
When you have a decimal (such as 5.6), each number is a representation of a place. (Tenths, Hundreds, Thousands, etc.)
So in a random example 0.123 - 1 is the tenths, 2 is the hundreds, 3 is the thousands.
Multiplying any 2 digit (or less) decimal by 100 (or more) changes the location of that decimal by moving it to the right by the given "places" (1 for 10, 2 for hundred, 3 for thousands).
So 5.6 becomes 560.0 when multiplied by 100. We have moved the decimal two places to the right, because we are multiplying by the hundreds place.
As another example, let's say we had a decimal of 0.561. We are now in the "thousands" place in decimals, so the only way to get rid of the decimal is to multiple the number by 1000 (or more), because the decimal is long enough to hit the "thousands" place.
At the same time, with the decimal "5.6", we could multiply it by 10, because it only has one decimal and therefore is in the "tenths" place. That makes it "56" instead of "560".
So why do we have to use 100 instead of 10? Because the question requires 1 answer for the entire equation, and in the equation is the number "0.12", which hits the "hundreds" place.
Why 10 or 100 and not 11 or 134 (ie any number greater than these 10 and 100)? Two reasons - first it is the lowest number possible to reach our goal. Second is because using round numbers by 10s preserves the original decimal easily for our eyes.
If we did 5.6 x 134 (for example) we end up with 750.4. If I handed that to Bob, he'd have to do some complicated math to get the answer back to the original equation. Our math would not be incorrect (as long as we used 134 for every number in the equation), but it would be annoying! And it makes us more at risk to mess up our math!
Let me know if I can clarify anything further.
Answer:
100
Step-by-step explanation:
To Find: Which number can each term of the equation be multiplied by to eliminate the decimals before solving? 5.6j – 0.12 = 4 + 1.1j
Solution :
5.6j – 0.12 = 4 + 1.1j
Since we can see that in the given equation the constants and the coefficients are in decimals
All the number are in decimal up to tenth place except 0.12
0.12 is in decimal up to hundredth place
So if we want to eliminate the decimal in 0.12. so we need to multiply it with 100
So, the given equation must be multiplied by 100 to make it decimal free
⇒100*( 5.6j – 0.12 = 4 + 1.1j)
⇒56j-12=4+11j
Thus Option D is correct : 100
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