6x – 6y = 0 3x + y =- 12
1 answer:
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Quotient is basically the answer you get when you divide two numbers.
For example, the quotient of 6 and 3 is 2 because when I divide 6 by 3, I get 2.
Quotient of x and 3 =
Eight less than quotient of x and 3 is
.
Hence, the answer is C.
For example, the quotient of 6 and 3 is 2 because when I divide 6 by 3, I get 2.
Quotient of x and 3 =
Eight less than quotient of x and 3 is
Hence, the answer is C.
Given:
For every 3 seeds Lara planted, only 2 seeds grew.
Altogether, 12 seeds grew.
To find:
The number of total seeds planted by Lara.
Solution:
It is given that for every 3 seeds Lara planted, only 2 seeds grew. It means the ratio of total seeds planted to seeds grew is 3:2.
Altogether, 12 seeds grew.
Let x be the number of total seeds planted by Lara. Then,
Therefore, the total number of seeds planter by Lara is 18.
Note that 6% converted to a decimal number is 6/100=0.06. Also note that 6% of a certain quantity x is 0.06x.
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
.
Similarly we can check that in the year 1987 the worker earned
, which makes the pattern clear.
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
.
Factorizing, we have
![=10,500[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]=10,500\cdot[1+1.06+(1.06)^2+(1.06)^3+...+(1.06)^{26}]](https://tex.z-dn.net/?f=%3D10%2C500%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D%3D10%2C500%5Ccdot%5B1%2B1.06%2B%281.06%29%5E2%2B%281.06%29%5E3%2B...%2B%281.06%29%5E%7B26%7D%5D)
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
(where a is the first term and r is the common ratio) we have:
.
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
.
Answer: D
Here is how much the worker earned each year:
In the year 1985 the worker earned <span>$10,500.
</span>In the year 1986 the worker earned $10,500 + 0.06($10,500). Factorizing $10,500, we can write this sum as:
$10,500(1+0.06).
In the year 1987 the worker earned
$10,500(1+0.06) + 0.06[$10,500(1+0.06)].
Now we can factorize $10,500(1+0.06) and write the earnings as:
$10,500(1+0.06) [1+0.06]=
Similarly we can check that in the year 1987 the worker earned
We can count that from the year 1985 to 1987 we had 2+1 salaries, so from 1985 to 2010 there are 2010-1985+1=26 salaries. This means that the total paid salaries are:
Factorizing, we have
We recognize the sum as the geometric sum with first term 1 and common ratio 1.06, applying the formula
Finally, multiplying 10,500 by 59.17 we have 621.285 ($).
The answer we found is very close to D. The difference can be explained by the accuracy of the values used in calculation, most important, in calculating
Answer: D