Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
1) A helium balloon is rising according to the function H(t) = -7x^2 + 5x - 11 where h(t) is the height in meters after t secon...Asked by Dana
1) A helium balloon is rising according to the function H(t) = -7x^2 + 5x - 11 where h(t) is the height in meters after t seconds. Determine the average rates of increase of height from t=7 to t=23 seconds.
I tried
h(23) - h(7)
------------
23-7
= * in this step i subbed it into the equation.
= -3599 + 319
-------------
16
It seems odd...
2) What is the end behaviour of
f(x) = 2-3x+4x^2
-----------
8x^2+5x-11
I understand end behaviour..but can someone help rearrange the first equation?
3) 2^2x - 4 (2^x) + 3 = 0
Solve for x.
This one I don't get at all.m
I tried
h(23) - h(7)
------------
23-7
= * in this step i subbed it into the equation.
= -3599 + 319
-------------
16
It seems odd...
2) What is the end behaviour of
f(x) = 2-3x+4x^2
-----------
8x^2+5x-11
I understand end behaviour..but can someone help rearrange the first equation?
3) 2^2x - 4 (2^x) + 3 = 0
Solve for x.
This one I don't get at all.m
Answers
Answered by
Damon
It is odd all right. First of all if it is h(t), why is it written with x and no t on the right? The whole thing is a typo.
The -x^2 term means it will sink, not rise, as x gets large. That is why your average speed comes out negative.
2-3x+4x^2
-----------
8x^2+5x-11
4 x^2 -3 x + 2
------------------
8 x^2 + 5 x -11
as x gets big, the x^2 terms are much bigger than the terms in x and the constants so
4 x^2
------ = 1/2
8 x^2
2^2x - 4 (2^x) + 3 = 0
The trick is to see that:
2^2x = (2^x)^2
so say y = 2^x
and we have
y^2 - 4 y +3 = 0
(y-3)(y-1)=0
y = 3 or y = 1
so 2^x = 3 or 2^x = 1
x log 2 = log 3 or x log 2 = log 1
x =.477/.301 or x = 0
The -x^2 term means it will sink, not rise, as x gets large. That is why your average speed comes out negative.
2-3x+4x^2
-----------
8x^2+5x-11
4 x^2 -3 x + 2
------------------
8 x^2 + 5 x -11
as x gets big, the x^2 terms are much bigger than the terms in x and the constants so
4 x^2
------ = 1/2
8 x^2
2^2x - 4 (2^x) + 3 = 0
The trick is to see that:
2^2x = (2^x)^2
so say y = 2^x
and we have
y^2 - 4 y +3 = 0
(y-3)(y-1)=0
y = 3 or y = 1
so 2^x = 3 or 2^x = 1
x log 2 = log 3 or x log 2 = log 1
x =.477/.301 or x = 0
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.