Solve the following system using elimination:
{-2 x + 2 y + 3 z = 0 | (equation 1)
{-2 x - y + z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Subtract equation 1 from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x - 3 y - 2 z = -3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Multiply equation 2 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{2 x + 3 y + 3 z = 5 | (equation 3)
Add equation 1 to equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+3 y + 2 z = 3 | (equation 2)
{0 x+5 y + 6 z = 5 | (equation 3)
Swap equation 2 with equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+3 y + 2 z = 3 | (equation 3)
Subtract 3/5 × (equation 2) from equation 3:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - (8 z)/5 = 0 | (equation 3)
Multiply equation 3 by 5/8:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y - z = 0 | (equation 3)
Multiply equation 3 by -1:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y + 6 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+5 y+0 z = 5 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 2 by 5:
{-(2 x) + 2 y + 3 z = 0 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Subtract 2 × (equation 2) from equation 1:
{-(2 x) + 0 y+3 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
v0 x+0 y+z = 0 | (equation 3)
Subtract 3 × (equation 3) from equation 1:
{-(2 x)+0 y+0 z = -2 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Divide equation 1 by -2:
{x+0 y+0 z = 1 | (equation 1)
{0 x+y+0 z = 1 | (equation 2)
{0 x+0 y+z = 0 | (equation 3)
Collect results:
Answer: {x = 1, y = 1, z = 0
- you will need 2 busses to only transport the boys.
- Mark is at (1 + 5/6) miles of his house.
<h3>How many buses would it take to carry only the boys?</h3>
We know that there are (3 + 1/2) groups, such that each group fill one bus.
2/5 of the students are boys, then the number of groups that we can make only with boys is:
(2/5)*(3 + 1/2) = 6/5 + 1/5 = 7/5 = 5/5 + 2/5 = 1 + 2/5
Then you can make one and a little less than a half of a group, which means that you need 1 and 2/5 of a buss to transport the boys, rounding that to the a whole number, you will need 2 busses to only transport the boys.
<h3>How far is Mark from his house?</h3>
The original distance is:
D = (2 + 3/4) miles.
But Mark only covers 2/3 of that distance, then we have:
d = (2/3)*D = (2/3)*(2 + 3/4) miles = (4/3 + 2/4) miles
d = (4/3 + 1/2) miles = (8/6 + 3/6) miles = (1 + 5/6) miles
Mark is at (1 + 5/6) miles of his house.
If you want to learn more about mixed numbers:
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Step-by-step explanation:
- She decides to put her money in the bank in an account that has a 10% interest rate that is compounded continuously.
Part a) What type of exponential model is Natalie’s situation?
Answer:
As Natalie's situation implies
- continuous compounding. So, instead of computing interest on a finite number of time periods, for instance monthly or yearly, continuous compounding computes interest assuming constant compounding over an infinite number of periods.
So, it requires the more generalized version of the principal calculation formula such as:
![P\left(t\right)=P_0\times \left[1+\left(i\:/\:n\right)\right]^{\left(n\:\times \:\:t\right)}](https://tex.z-dn.net/?f=P%5Cleft%28t%5Cright%29%3DP_0%5Ctimes%20%5Cleft%5B1%2B%5Cleft%28i%5C%3A%2F%5C%3An%5Cright%29%5Cright%5D%5E%7B%5Cleft%28n%5C%3A%5Ctimes%20%5C%3A%5C%3At%5Cright%29%7D)
or
![P\left(t\right)=P_0\times \left[1+\left(\frac{i}{n}\:\right)\right]^{\left(n\:\times \:\:t\right)}](https://tex.z-dn.net/?f=P%5Cleft%28t%5Cright%29%3DP_0%5Ctimes%20%5Cleft%5B1%2B%5Cleft%28%5Cfrac%7Bi%7D%7Bn%7D%5C%3A%5Cright%29%5Cright%5D%5E%7B%5Cleft%28n%5C%3A%5Ctimes%20%5C%3A%5C%3At%5Cright%29%7D)
Here,
= interest rate
= number of compounding periods
= time period in years
Part b) Write the model equation for Natalie’s situation?
For continuous compounding the number of compounding periods, 
, becomes infinitely large.
Therefore, the formula as we discussed above would become:
Part c) How much money will Natalie have after 2 years?
Using the formula
$₂ 
$
So, Natalie will have 
$ after 2 years.
Part d) How much money will Natalie have after 2 years?
Using the formula
$₁₀ 
$
So, Natalie will have 
$ after 10 years.
Keywords: word problem, interest
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