Answer:
Step-by-step explanation:
FV =<u> p (1+i)^n -1</u>
i
pv = 700,000
i = .075/12 = .00625
n = (66 - 15)* 12 = 612
700,000 = P (( 1 + .00625)^ 612 -1 /.00625
4375 = P (1.00625)^612 -1)
P = $98.77
First we must construct an equation to model the problem. (In this case we will use an inequality instead) This is what I came up with:
450.20+0.15s>=600.10
This equation shows how if her base earnings ($450.20) are added to 15% of her sales, represented by s, then the total will be greater than or equal to $600.10
Next, we simply solve for s. (steps shown below)
1) 450.20+0.15s>=600.10 (simply restating the inequality)
2) 0.15s>=149.90 (here I isolated the variable)
3)0.15s/0.15>=149.90/0.15 (Finally I solve for s by dividing both sides by 0.15, this will isolate s on the left and leave the answer on the right)
4) s>=999.33... (here I found the total sales the salesperson would need to reach his/her goal of earning a minimum of $600.10; the 3's after the decimal are repeating so in the next step I will round up to the nearest hundredth (b/c this is what money is rounded to and if I round down he/she would make less than her goal. This means i must round up.))
5) s>=999.34 (simple rounding; once again I rounded up b/c rounding down would slightly bring the total earnings to less than the goal)
<u>Therefore, the salesperson would need his/her sales to be $999.34 in order for his/her total earnings for the week to be at least $600.10</u> (greater than or equal to $600.10)
<u>Hope this helped!</u>
Answer:
X <u>></u> 7
Step-by-step explanation:
X <u>></u> 7
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
Now, the system of equations in matrix form will be:
![\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%26%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%26%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26%5C%5Cy%26%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%26%5C%5C7%26%5Cend%7Barray%7D%5Cright%5D%20)
Next, we are going to find the determinant:
![D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}](https://tex.z-dn.net/?f=D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C%20%5Cfrac%7B1%7D%7B65%7D%20%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%281%29%28%20%5Cfrac%7B1%7D%7B25%7D%29%20-%20%281%29%28%20%5Cfrac%7B1%7D%7B65%7D%20%29%3D%20%5Cfrac%7B8%7D%7B325%7D%20)
Next, we are going to find the determinant of x:
![D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8](https://tex.z-dn.net/?f=%20D_%7Bx%7D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D375%261%5C%5C7%26%20%5Cfrac%7B1%7D%7B25%7D%20%5Cend%7Barray%7D%5Cright%5D%20%3D%20%28375%29%28%20%5Cfrac%7B1%7D%7B25%7D%20%29-%281%29%287%29%3D8)
Now, we can find x:
Now that we know the value of x, we can find y:
Remember that time equals distance over velocity; therefore, the time on the highway will be:
An the time on the city will be:
We can conclude that the bus was five hours on the highway and two hours in the city.
Answer:
Option A is correct.
Step-by-step explanation:
The height of the rectangular pack is h1 = 4r
The height of the parallelogram = 2r + the height of equilateral triangles with a side of 2r
The height of equilateral triangle with a side of 2r = (side)/2*√3
=(2r)/2 *√3 = r√3
The height of the parallelogram h2 = 2r+r√3 = r(2+√3)
So,the ratio of h1 to h2 is :
4r : r(2 +√3)
Cancelling r we get
Hence, the ratio is 4 : 2 +√3
