With two employee shifts, each will work 12 hours (2 x 12)
With three employee shifts, each will work 8 hours (3 x 8)
With four employee shifts, each will work 6 hours (4 x 6)
With six employee shifts, each will work 4 hours ( 6 x 4)
With twelve employees shifts, each will work 2 hours ( 12 x 2)
Answer:
2) x = 6 y = 7 3) x = -5 y= 2
Step-by-step explanation:
2) 10x - 8y = 4
2(-5x + 3y = -9)
-10x + 6y = -18
10x - 8y = 4 Add both equations together
-2y = -14
y = 7 Divide both sides by -2
10(x) - 8(7) = 4 Plug in 7 for the y to solve for x
10x - 56 = 4
+ 56 + 56 Add 56 to both sides
10x = 60
x = 6 Divide both sides by 10
3) Where the two lines intersect, that is the solution to x and y.
The lines intersect at (-5,2). x = -5 y = 2
. Let me do my best.
<span>$187,500 is cost of house. </span>
<span>20%, or $37,500 is the down payment. </span>
<span>The loan amount would be $187,500 - $37,500 = $150,000. </span>
<span>If we assume the annual rate of the loan is 4.65% </span>
<span>Then the monthly rate would be 4.65%/12 = 0.3875% </span>
<span>If the loan is $150,000, the interest is 0.3875% </span>
<span>The interst for the first month is $150,000 * 0.3875% = $581.25. </span>
<span>You stated that their payment is $1,575. </span>
<span>So the amount that pays off the loan is $1,575 - $581.25 = $993.75. </span>
<span>At the end of the month, they owe $150,000 - $993.75 = $149,006.25 </span>
<span>For the second month, the amount of the payment that goes towards interst is </span>
<span>$149,006.25 * 0.3875% = $577.40. and the amount that goes towards the loan is $997.60. </span>
<span>At the end of the second month they owe $148,008.65. </span>
<span>Regarding realized income, we recommend a monthly loan payment not to exceed 28% of the monthly income. So if a payment of $1,575 is 28% of Gross, then the math is : $1,575 = 0.28*Gross. </span>
<span>Gross = $5,625 monthly. </span>
<span>About $67,500 annually. </span>
<span>About $33.75 an hour.</span>
Exponentail thingies
easy, look at all them them, see that they have 5 in common?
rremember how esay it was to factor
ax^2+bx+c=0
now we have
5^(2x)-6(5^x)+5=0
remember that 5^(2x)=(5^2)^x or (5^x)^2
in other words, we can rewrite it as
1(5^x)^2-6(5^x)+5=0
if yo want, replace 5^x with a and factor
1a^2-6a+5=0
(a-1)(a-5)=0
a=5^x
(5^x-1)(5^x-5)=0
set each to zero
5^x-1=0
5^x=1
take the log₅ of both sides
x=log₅1
5^x-4=0
5^x=4
take the log₅ of both sides
x=log₅4
x=log₅1 and/or log₅4
second quesiton
same thing
1(2^x)-10(2^x)+16=0
factor
(2^x-8)(2^x-2)=0
set each to zero
2^x-8=9
2^x=8
x=3
2^x-2=0
2^x=2
x=1
x=3 or 1
first one
x=log₅1 and/or log₅4
second one
x=1 and/or 3
3 1/3, 7-3 2/3=3 1/3, use absolute value