I think the answer will be t=2/3 or t= -2/3
Answer:
x=-5
Step-by-step explanation:
I did a for #42
2(x-5)=3x+5
simplify
2x-10=3x+5
subtract 2x from both sides
-10=3x+5
subtract 5 from both sides
-15=3x
divide both sides by 3
-5=x
Answer:
Step-by-step explanation:
I'm sure you want your functions to appear as perfectly formed as possible so that others can help you. f(x) = 4(2)x should be written with the " ^ " sign to denote exponentation: f(x) = 4(2)^x
f(b) - f(a)
The formula for "average rate of change" is a.r.c. = --------------
b - a
change in function value
This is equivalent to ---------------------------------------
change in x value
For Section A: x changes from 1 to 2 and the function changes from 4(2)^1 to 4(2)^2: 8 to 16. Thus, "change in function value" is 8 for a 1-unit change in x from 1 to 2. Thus, in this Section, the a.r.c. is:
8
------ = 8 units (Section A)
1
Section B: x changes from 3 to 4, a net change of 1 unit: f(x) changes from
4(2)^3 to 4(2)^4, or 32 to 256, a net change of 224 units. Thus, the a.r.c. is
224 units
----------------- = 224 units (Section B)
1 unit
The a.r.c for Section B is 28 times greater than the a.r.c. for Section A.
This change in outcome is so great because the function f(x) is an exponential function; as x increases in unit steps, the function increases much faster (we say "exponentially").
Answer: -a2 - 2ab - b2 + 25 =
-1 • (a2 + 2ab + b2 - 25)
Final result :
-a2 - 2ab - b2 + 25
Step-by-step explanation:
Answer: C) 127, 152.4, 182.88, 219.456,...
Step-by-step explanation:
You can only find the sum of an infinite geometric sequence if it converges.
One criterion to see if the series converges is if:
aₙ < aₙ₋₁
This means that, as n increases, the value of the terms decreases.
This means that as n tends to infinity, aₙ tends to zero.
Then we only can find the sum of those series where the terms are decreasing.
in A, B and D the terms are decreasing, then we can find the sum of those 3 series.
Now in the case of C, the terms are increasing, then we can not find the sum of that series.
