Answer:
If your asking us to just combine like terms it would be -3b^2-2b^2+1.4b+4c+c
but if your asking us to answer the question it would be -5b^2+1.4b+5c
Hope I helped :)
Answer:
$6500
Step-by-step explanation
x = amount ($) loaned at 6%
then
18000-x = amount ($) loaned at 14%
.06x + .14(18000-x) = 2000
.06x + 2520-.14x = 2000
2520-.08x = 2000
-.08x = -520
x = $6,500
1. so it has to have every x is one y and every y has one x
graph them
we see that the only ones that are one to one are the first one, the 2nd one and the 4th one
2.
solve for x and replace with f⁻¹(x) and y with x
minus 8 and cube both sides
(y-8)³=x-2
add 2
(y-8)³+2=x
replace
f⁻¹(x)=(x-8)³+2
first one
3.
this is a one to one function because theer is no vertical line that could intersect the graph more than once
ah, I see, you're program has a different one to one definition, no horizontal or vertical line can cross, lemme edit te first question again
answer is 2nd option
4. a neat trick is this:
the domain of f(x) is the range of f⁻¹(x)
the range of f(x) is the domain of f⁻¹(x)
the domain of f(x)=1/(3x+2)
hmm, can't divide by 0 so set denomenator to 0
3x+2=0
3x=-2
x=-2/3
domain is all real numbers except -2/3
(-infinity,-2/3)U(-2/3,infinity)
that's the range of f⁻¹(x)
range
ok, hmm, range is from positive initny to 0 then from 0 to positive inifnty, not including 0
so the domain of f⁻¹(x) is (-∞,0)U(0,∞)
range of f⁻¹(x) is (-∞,-2/3)U(-2/3,∞)
first option
Answer:
410.32
Step-by-step explanation:
Given that the initial quantity, Q= 6200
Decay rate, r = 5.5% per month
So, the value of quantity after 1 month, 
The value of quantity after 2 months, 
From equation (i)
The value of quantity after 3 months, 
From equation (ii)
Similarly, the value of quantity after n months,
As 4 years = 48 months, so puttion n=48 to get the value of quantity after 4 years, we have,
Putting Q=6200 and r=5.5%=0.055, we have
Hence, the value of quantity after 4 years is 410.32.
