Question
Cable company A charges $45 a month for cable plus a $18 installation fee. Cable B charges $39 a month for cable plus a $30 fee for installation. Which inequality can be used to find out when the monthly cost for Cable company A is less than Cable company B?
a)39x+30<45x+18
b) 45x+18>39x+30
c) 45x+18<39x+30
d)39x+18<45x+30
Answer:
c) 45x+18<39x+30
Step-by-step explanation:
Step 1
We have to find the Algebraic expressions for the cable companies
Cable company A charges $45 a month for cable plus a $18 installation fee.
Let the number of months be represented as x
Hence, this is represented as:
$45 × x + $18
= 45x + 18
Cable B charges $39 a month for cable plus a $30 fee for installation.
Let the number of months be represented as x
Hence, this is represented as:
$39 × x + $30
= 39x + 30
The inequality that can be used to find out when the monthly cost for Cable company A is less than Cable company B?
= Cable company A < Cable company B
= 45x + 18 < 39x + 30
Therefore option c is correct
Answer:
the dimensions of the most economical shed are height = 10 ft and side 5 ft
Step-by-step explanation:
Given data
volume = 250 cubic feet
base costs = $4 per square foot
material for the roof costs = $6 per square foot
material for the sides costs = $2.50 per square foot
to find out
the dimensions of the most economical shed
solution
let us consider length of side x and height is h
so we can say x²h = 250
and h = 250 / x²
now cost of material = cost of base + cost top + cost 4 side
cost = x²(4) + x²(6) + 4xh (2.5)
cost = 10 x² + 10xh
put here h = 250 / x²
cost = 10 x² + 10x (250/ x² )
cost = 10 x² + (2500/ x )
differentiate and we get
c' = 20 x - 2500 / x²
put c' = 0 solve x
20 x - 2500 / x² = 0
x = 5
so we say one side is 5 ft base
and height is h = 250 / x²
h = 250 / 5²
height = 10 ft
Answer:
The area of the searched region is 
Step-by-step explanation:
If you want to find the area of a region bounded by functions f(x) and G(x) between two limits (a,b), you have to do a double integral. you must first know which of the functions is greater than the other for the entire domain.
In this case, for 0<x<1, f(x)<g(x)
while for 1<x, g(x)<f(x).
Therefore if our domain is all real numbers superior to 0 (where the limit 0<a<1 and 1<b), we have to do 2 integrals:
A=A(a<x<1)+A(1<x<b)
Step-by-step explanation:
yxbxbxbxbxbxdbdbdbdnxjcjkddkekekennendif
It seems there are not fractions between 3/5 and 4/5, but there are so many
for example, if you make denominator to 10 for both fractions, you will get 6/10 and 8/10
you can say 7/10 is in between 3/5 and 4/5
also, if you make denominator to 100 for both fractions, you will get 60/100 and 80/100
you can get 20 fractions between 3/5 and 4/5
