Answer:
Option a. $300
Step-by-step explanation:
Let
x ----> the cost of the rent monthly per person
we know that
The number of persons is 3 (Sal and two friends)
The cost of the security deposit per person is equal to
If Sal paid a total of $1,000 to move in
then
That means that the cost per person of two months of rent (first and last month) plus $400 (security deposit per person) must be equal to $1,000
solve for x
Answer:
Step-by-step explanation:
<u>Initial balance:</u>
<u>Adding deposits: </u>
- $85.62 + $86.90 + $6.74 = $179.26
<u>Subtracting withdrawals:</u>
- $179.26 - ($73.59 + $7.81 + $115.75) = - $17.89
<u>End balance:</u>
Answer:
a = 22
b = 31
c = 13
Step-by-step explanation:
The sum is the same in each row, column, and diagonal.
One of the diagonals is already complete. The sum is:
16 + 25 + 34 = 75
So the first row adds up to 75:
a + 37 + 16 = 75
a = 22
The second row adds up to 75:
19 + 25 + b = 75
b = 31
And the third row adds up to 75:
34 + c + 28 = 75
c = 13
We can check our answer by finding the sum of each column and the other diagonal.
22 + 19 + 34 = 75
37 + 25 + 13 = 75
16 + 31 + 28 = 75
22 + 25 + 28 = 75
Answer:
$20
Step-by-step explanation:
7 x __=70
2 x ____=___
7 x 10 = 70
2 x 10
=20
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27). But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing. The first derivative of the function is
. Set the derivative equal to 0 and factor to find the critical numbers.
, so x = -3 and x = 1/3. We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing. From there we will look at our points to determine which fall into the "decreasing" category. Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞. In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval. In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval. In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval. In order to determine where our points in question fall, look to the x value. The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12). The point (-3, -18) is already a min value.