Answer:
The number of students that bring their lunches is 12
Step-by-step explanation:
Let
x -----> the number of students that bring their lunches
y -----> the total number of students in a class
we know that
The number of students that bring their lunches divided by the total number of students in a class must be equal to 3/8
-----> equation A
-----> equation B
substitute the value of y in equation A and solve for x
therefore
The number of students that bring their lunches is 12
Answer:
You shoud invest $0.204, it is an small amount each month.
The formula to calculate the monthly saving is
![PP= \frac{A(APR/n)}{(1+(APR/n))^{nY}-1}](https://tex.z-dn.net/?f=PP%3D%20%5Cfrac%7BA%28APR%2Fn%29%7D%7B%281%2B%28APR%2Fn%29%29%5E%7BnY%7D-1%7D)
with PP: periodic payments, A: amount desired, n: number of payments in one year, Y: number of years.
In this case A = 88, n= 12 (the payments are monthly), Y=18, APR=7%.
Replacing in the formula you obtain that ![PP \approx 0.204](https://tex.z-dn.net/?f=PP%20%5Capprox%200.204)
Answer:
7/14 and 6/14
Step-by-step explanation:
we are trying to find the same denominator for the 2 fractons, so 1/2 times 2 to both sides would be 7/14. and 3/7 times 2 to both sides would be 6/14. (same denominator)
Answer:
Step-by-step explanation:
17 : i 12x^2(4x−3) ,
ii : 5x(x−3y) ,
iii: 5xy^2z(3x^2−5z^2)
18) since the two polynomial are equal then :
2x^3+ax^2+3x-5 = x^3+x^2-2x+a same remainder x-2 then x=2
2(2)^3+a(2)^2+3(2)-5=(2)^3+(2)^2-2(2)+a solve for a
-3a=9
a=-3
18 ) x^3+y^3 -125+15xy
x^3+y^3-(5)^3-3(x)(y)(-5) then factorize x^3+y^3
(x+y)(x^2-xy+y^2)- (5)(5)^2-3(x)(y)(-5) common factor x+y-5
(x+y-5)(x^2-xy+y^2-5^2-3xy given x+y=5
5-5( x^2-xy+y^2-25-3xy) = 0
the value of the expression equal to zero