Posted by Dana on Wednesday, January 23, 2008 at 12:37am.
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)

237
= * in this step i subbed it into the equation.
= 3599 + 319

16
It seems odd...
2) What is the end behaviour of
f(x) = 23x+4x^2

8x^2+5x11
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

Math  Damon, Wednesday, January 23, 2008 at 6:01am
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.
23x+4x^2

8x^2+5x11
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
(y3)(y1)=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
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